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Elementary Number Theory
WISB321
=
F.Beukers
Department of Mathematics
2012
UU
ELEMENTARY NUMBER THEORY
Frits Beukers
Fall semester 2013
Contents
1 Integers and the Euclidean algorithm
1.1 Integers . . . . . . . . . . . . . . . .
1.2 Greatest common divisors . . . . . .
1.3 Euclidean algorithm for Z . . . . . .
1.4 Fundamental theorem of arithmetic .
1.5 Exercises . . . . . . . . . . . . . . . .
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2 Arithmetic functions
2.1 Definitions, examples . . . . . . . . . . . . . . . . . . . . . . . . .
2.2 Convolution, Möbius inversion . . . . . . . . . . . . . . . . . . . .
2.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
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3 Residue classes
3.1 Basic properties . . . . . . .
3.2 Chinese remainder theorem
3.3 Invertible residue classes . .
3.4 Periodic decimal expansions
3.5 Exercises . . . . . . . . . . .
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4 Primality and factorisation
4.1 Prime tests and compositeness tests . .
4.2 A polynomial time primality test . . .
4.3 Factorisation methods . . . . . . . . .
4.4 The quadratic sieve . . . . . . . . . . .
4.5 Cryptosystems, zero-knowledge proofs .
5 Quadratic reciprocity
5.1 The Legendre symbol . . .
5.2 Quadratic reciprocity . . .
5.3 A group theoretic proof . .
5.4 Applications . . . . . . . .
5.5 Jacobi symbols, computing
5.6 Class numbers . . . . . . .
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square
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2
CONTENTS
5.7
Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
6 Dirichlet characters and Gauss sums
6.1 Characters . . . . . . . . . . . . . . .
6.2 Gauss sums, Jacobi sums . . . . . . .
6.3 Applications . . . . . . . . . . . . . .
6.4 Exercises . . . . . . . . . . . . . . . .
7 Sums of squares, Waring’s problem
7.1 Sums of two squares . . . . . . . .
7.2 Sums of more than two squares . .
7.3 The 15-theorem . . . . . . . . . . .
7.4 Waring’s problem . . . . . . . . . .
7.5 Exercises . . . . . . . . . . . . . . .
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8 Continued fractions
8.1 Introduction . . . . . . . . . . . . . . . . . .
8.2 Continued fractions for quadratic irrationals
8.3 Pell’s equation . . . . . . . . . . . . . . . . .
8.4 Archimedes’s Cattle Problem . . . . . . . .
8.5 Cornacchia’s algorithm . . . . . . . . . . . .
8.6 Exercises . . . . . . . . . . . . . . . . . . . .
9 Diophantine equations
9.1 General remarks . . . . . .
9.2 Pythagorean triplets . . .
9.3 Fermat’s equation . . . . .
9.4 Mordell’s equation . . . .
9.5 The ‘abc’-conjecture . . .
9.6 The equation xp + y q = z r
9.7 Mordell’s conjecture . . .
9.8 Exercises . . . . . . . . . .
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10 Prime numbers
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10.1 Introductory remarks . . . . . . . . . . . . . . . . . . . . . . . . . 107
10.2 Elementary methods . . . . . . . . . . . . . . . . . . . . . . . . . 111
10.3 Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 114
11 Irrationality and transcendence
11.1 Irrationality . . . . . . . . . . .
11.2 Transcendence . . . . . . . . . .
11.3 Irrationality of ζ(3) . . . . . . .
11.4 Exercises . . . . . . . . . . . . .
F.Beukers, Elementary Number Theory
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CONTENTS
3
12 Solutions to selected problems
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13 Appendix: Elementary algebra
13.1 Finite abelian groups . . . . .
13.2 Euclidean domains . . . . . .
13.3 Gaussian integers . . . . . . .
13.4 Quaternion integers . . . . . .
13.5 Polynomials . . . . . . . . . .
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F.Beukers, Elementary Number Theory
Chapter 1
Integers and the Euclidean
algorithm
1.1
Integers
Roughly speaking, number theory is the mathematics of the integers. In any
systematic treatment of the integers we would have to start with the so-called
Peano-axioms for the natural numbers, define addition, multiplication and ordering on them and then deduce their elementary properties such as the commutative, associatative and distributive properties. However, because most students
are very familiar with the usual rules of manipulation of integers, we prefer to
shortcut this axiomatic approach. Instead we simply formulate the basic rules
which form the basis of our course. After all, we like to get as quickly to the
parts which make number theory such a beautiful branch of mathematics.
We start with the natural numbers
N : 1, 2, 3, 4, 5, . . .
On N we have an addition (+) and multiplication (× or ·) law and a well-ordering
(>, <, ≥, ≤). By a well-ordering we mean that
1. For any distinct a, b ∈ N we have either a > b or a < b.
2. From a < b and b < c follows a < c
3. There is a smallest element, namely 1. So a ≥ 1 for all a ∈ N.
We shall assume that we are all familiar with the usual rules of addition and
multiplication.
1. For all a, b ∈ N: a + b = b + a and ab = ba (commutativity of addition and
multiplication).
4
1.1. INTEGERS
5
2. For all a, b, c ∈ N: (a + b) + c = a + (b + c) and (ab)c = a(bc) (associativity
of addition and multiplication)
3. For all a, b, c ∈ N: a(b + c) = ab + ac (distributive law).
4. For all a ∈ N: 1 · a = a.
5. For all a ∈ N: a + 1 > a.
6. For all a, b, c ∈ N: b > c ⇒ a + b > a + c and b ≥ c ⇒ ab ≥ ac.
7. For all a, b ∈ N: a > b ⇒ there exists c ∈ N such that a = b + c.
We shall also use the following fact.
Theorem 1.1.1 Every non-empty subset of N has a smallest element.
Then there is the principle of induction.
Theorem 1.1.2 Let S ⊂ N and suppose that
1. 1 ∈ S
2. For all a ∈ N: a ∈ S ⇒ a + 1 ∈ S
Then S = N.
Theorem 1.1.2 follows from Theorem 1.1.1 in the following way. Let S be as
in Theorem 1.1.2 and consider the complement S c . This set is either empty, in
which case Theorem 1.1.2 is proven, or S c is non-empty. Let us assume the latter.
Theorem 1.1.1 states that S c has a smallest element, which we denote by a. If
a = 1, then a ̸∈ S, violating the first condition of Theorem 1.1.2. If a > 1 then
a − 1 ̸∈ S c . Hence a − 1 ∈ S and a ̸∈ S, violating the second condition. We
conclude that S c is empty, hence S = N.
We call a subset S ⊂ N finite if there exists m ∈ N such that s < m for all s ∈ S.
There are two concepts which partially invert addition and multiplication.
1. Subtraction Let a, b ∈ N and a > b. Then there exists a unique c ∈ N
such that a = b + c. We call c the difference between a and b. Notation;
a − b.
2. Divisibility We say that the natural number b divides a if there exists
c ∈ N such that a = bc. Notation: b|a, and b is called a divisor of a.
There are many well-known, almost obvious, properties which are not mentioned
in the above rules, but which nevertheless follow in a more or less straightforward
way. As an exercise you might try to prove the following properties.
F.Beukers, Elementary Number Theory
6
CHAPTER 1. INTEGERS AND THE EUCLIDEAN ALGORITHM
1. For all a, b, c ∈ N: a + b = a + c ⇒ b = c
2. For all a, b, c ∈ N: ab = ac ⇒ b = c.
3. For all a, b, d ∈ N: d|a, d|b ⇒ d|(a + b)
4. Any a ∈ N has finitely many divisors.
5. Any finite set of natural numbers has a biggest element.
Although divison of one number by another usually fails we do have the concept
of division with remainder.
Theorem 1.1.3 (Euclid) Let a, b ∈ N with a > b. Then either b|a or there
exist q, r ∈ N such that
a = bq + r,
r < b.
Moreover, q, r are uniquely determined by these (in)equalities.
Proof. Suppose b does not divide a. Consider all multiples of b which are less
than a. This is a non-empty set, since b < a. Choose the largest multiple and
call it bq. Then clearly a − bq < b. Conversely, if we have a multiple qb such that
a − bq < b then qb is the largest b-multiple < a. Our theorem follows by taking
r = a − bq.
2
Another important concept in the natural numbers are prime numbers. These
are natural numbers p > 1 that have only the trivial divisors 1, p. Here are the
first few:
2, 3, 5, 7, 11, 13, 17, 19, 23, 29, 31, . . .
Most of us have heard about them at a very early age. We also learnt that there
are infinitely many of them and that every integer can be written in a unique way
as a product of primes. These are properties that are not mentioned in our rules.
So one has to prove them, which turns out to be not entirely trivial. This is the
beginning of number theory and we will take these proofs up in this chapter.
In the history of arithmetic the number 0 was introduced after the natural
numbers as the symbol with properties 0 · a = 0 for all a and a + 0 = a
for all a. Then came the negative numbers -1,-2,-3,. . . with the property that
−1 + 1 = 0, −2 + 2 = 0, . . .. Their rules of addition and multiplication are
uniquely determined if we insist that these rules obey the commutative, associate
and distributive laws of addition and multiplication. Including the infamous
”minus times minus is plus” which causes so many high school children great
headaches. Also in the history of mathematics we see that negative numbers
and their arithmetic were only generally accepted at a surprisingly late age, the
beginning of the 19th century.
F.Beukers, Elementary Number Theory
1.2. GREATEST COMMON DIVISORS
7
From now on we will assume that we have gone through all these formal introductions and we are ready to work with the set of integers Z, which consists of
the natural numbers, their opposites and the number 0.
The main role of Z is to have extended N to a system in which the operation
of subtraction is well-defined for any two elements. One may proceed further by
extending Z to a system in which also element (̸= 0) divides any other. The
smallest such system is well-known: Q, the set of rational numbers. At several
occasion they will also play an important role.
1.2
Greatest common divisors
Definition 1.2.1 Let a1 , . . . , an ∈ Z, not all zero. The greatest common divisor
of a1 , . . . , an is the largest natural number d which divides all ai
Notation: (a1 , . . . , an ) or gcd(a1 , . . . , an ).
Definition 1.2.2 Two numbers a, b ∈ Z, not both zero, are called relatively
prime if gcd(a, b) = 1.
Theorem 1.2.3 Let ai ∈ Z (i = 1, . . . , n) not all zero. Let d = gcd(a1 , . . . , an ).
Then there exist t1 , . . . , tn ∈ Z such that d = a1 t1 + · · · + an tn
Proof. Consider the set
S = {a1 x1 + · · · + an xn | x1 , · · · , xn ∈ Z}
and choose its smallest positive element. Call it s. We assert that d = s. First
note that every element of S is a multiple of s. Namely, choose x ∈ S arbitrary.
Then x − ls ∈ S for every l ∈ Z. In particular, x − [x/s]s ∈ S. Moreover,
0 ≤ x − [x/s]s < s. Because s is the smallest positive element in S , we have
necessarily x − [x/s]s = 0 and hence s|x. In particular, s divides ai ∈ S for every
i. So s is a common divisor of the ai and hence s ≤ d. On the other hand we
know that s = a1 t1 + · · · + an tn for suitable t1 , · · · , tn . From d|ai ∀i follows that
d|s. Hence d ≤ s. Thus we conclude d = s.
2
Corollary 1.2.4 Assume that the numbers a, b are not both zero and that a1 , . . . , an
are not all zero.
i. Every common divisor of a1 , · · · , an divides gcd(a1 , . . . , an ).
Proof: This follows from Theorem 1.2.3. There exist integers t1 , . . . , tn such
that gcd(a1 , . . . , an ) = a1 t1 + · · · + an tn . Hence every common divisor of
a1 , . . . , an divides their greatest common divisor.
F.Beukers, Elementary Number Theory
8
CHAPTER 1. INTEGERS AND THE EUCLIDEAN ALGORITHM
ii. Suppose gcd(a, b) = 1. Then a|bc ⇒ a|c.
Proof: ∃x, y ∈ Z : 1 = ax + by. So, c = acx + bcy. The terms on the right
are divisible by a and consequently, a|c.
iii. Let p be a prime. Then p|bc ⇒ p|b or p|c.
Proof: Suppose for example that p/|b, hence gcd(b, p) = 1. From (ii.) we
infer p|c.
iv. Let p be a prime and suppose p|a1 a2 · · · an . Then ∃i such that p|ai .
Proof Use (iii.) and induction on n.
v. Suppose gcd(a, b) = 1. then b|c and a|c ⇒ ab|c.
Proof: ∃x, y ∈ Z : 1 = ax + by. So, c = acx + bcy. Because both terms on
the right are divisible by ab we also have ab|c.
vi. gcd(a1 , . . . , an ) = gcd(a1 , . . . , an−2 , (an−1 , an )).
Proof: Every common divisor of an−1 and an is a divisor of (an−1 , an ) and
conversely (see i.). So, the sets a1 , . . . , an and a1 , . . . , an−2 , (an−1 , an ) have
the same common divisors. In particular they have the same gcd.
vii. gcd(a, b) = d ⇒ gcd(a/d, b/d) = 1.
Proof: There exist x, y ∈ Z such that ax+by = d. Hence (a/d)x+(b/d)y = 1
and so any common divisor of a/d and b/d divides 1, i.e. d = 1.
viii. gcd(a, b) = 1 ⇐⇒ there exist x, y ∈ Z : ax + by = 1.
1.3
Euclidean algorithm for Z
In this section we describe a classical but very efficient algorithm to determine the
gcd of two integers a, b and the linear combination of a, b which yields gcd(a, b).
First an example. Suppose we want to determine (654321,123456) . The basic
idea is that gcd(a, b) = gcd(a − rb, b) for all r ∈ Z. By repeatedly subtracting
the smallest term from the largest, we can see to it that the maximum of the
numbers between the gcd brackets decreases. In this way we get
=
=
=
=
=
=
gcd(654321, 123456) = gcd(654321 − 5 · 123456, 123456) = gcd(37041, 123456)
gcd(37041, 123456 − 3 · gcd(37041) = gcd(37041, 12333)
gcd(37041 − 3 · gcd(12333, 12333) = gcd(42, 12333)
gcd(42, 12333 − 293 · 42) = gcd(42, 27)
gcd(42 − 27, 27) = gcd(15, 27)
gcd(15, 27 − 15) = gcd(15, 12)
gcd(15 − 12, 12) = gcd(3, 12) = gcd(3, 0) = 3
F.Beukers, Elementary Number Theory
1.3. EUCLIDEAN ALGORITHM FOR Z
9
So we see that our greatest common divisor is 3. We also know that there exist
integers x, y such that 3 = 654321x + 123456y. To obtain such numbers we have
to work in a more schematic way where we have put a = 654321, b = 123456,
654321 = 5 · 123456 + 37041
123456 = 3 · 37041 + 12333
37041 = 3 · 12333 + 42
12333 = 293 · 42 + 27
42 = 1 · 27 + 15
27 = 1 · 15 + 12
15 = 1 · 12 + 3
12 = 4 · 3 + 0
654321 = 1a + 0b
123456 = 0a + 1b
37041 = 1a − 5b
12333 = −3a + 16b
42 = 10a − 53b
27 = −2933a + 15545b
15 = 2943a − 15598b
12 = −5876a + 31143b
3 = 8819a − 46741b
0 = −123456a + 654321b
In the left hand column we have rewritten the subtractions. In the righthand
column we have written all remainders as linear combinations of a = 654321 and
b = 123456.
In general, write r−1 = a en r0 = b and inductively determine ri+1 for i ≥ 0 by
ri+1 = ri−1 −[ri−1 /ri ]·ri until rk+1 = 0 for some k. Because r0 > r1 > r2 > · · · ≥ 0
such a k occurs. We claim that rk = gcd(a, b). This follows from the following
observation, gcd(ri , ri−1 ) = gcd(ri , ri−1 −[ri−1 /ri ]ri ) = gcd(ri , ri+1 ) = gcd(ri+1 , ri )
for all i. Hence gcd(b, a) = gcd(r0 , r−1 ) = gcd(r1 , r0 ) = · · · = gcd(rk , rk+1 ) =
gcd(rk , 0) = rk .
In our example we see in the right hand column a way to write 3 as a linear
combination of a = 654321 and b = 123456. The idea is to start with 654321 =
1 · a + 0 · b and b = 0 · a + 1 · b and combine these linearly as prescribed by
the euclidean algorithm. We obtain consecutively the ri as linear combination of
654321 and 123456 until
3 = 8819 · 654321 − 46741 · 123456.
In general, let ri (i ≥ −1) be as above and suppose rk = 0 and ri ̸= 0 (i < k).
Let
x−1 = 1, y−1 = 0, x0 = 0, y0 = 1
and inductively,
xi+1 = xi−1 − [ri−1 /ri ] · xi ,
yi+1 = yi−1 − [ri−1 /ri ] · yi
(i ≥ 0).
Then we can show by induction on i that ri = axi + byi for all i ≥ −1. In
particular we have gcd(a, b) = rk = axk + byk .
The proof of the termination of the euclidean algorithm is based on the fact that
the remainders r1 , r2 , . . . form a strictly decreasing sequence of positive numbers.
F.Beukers, Elementary Number Theory
10
CHAPTER 1. INTEGERS AND THE EUCLIDEAN ALGORITHM
In principle this implies that the number of steps required in the euclidean algorithm can be as large as the number a or b itself. This would be very bad if
the numbers a, b would have more than 15 digits, say. However, the very strong
point of the euclidean algorithm is that the number of steps required is very
small compared to the size of the starting numbers a, b. For example, 3100 − 1
and 2100 − 1 are numbers with 48 and 31 digits respectively. Nevertheless the
euclidean algorithm applied to them takes only 54 steps. Incidently, the gcd is
138875 in this case. All this is quantified in the following theorem.
Theorem 1.3.1 Let a, b ∈ N with a > b and apply the euclidean algorithm to
them. Use the same notations as above and suppose that rk is the last non-zero
remainder in the algorithm. Then
k<2
log a
.
log 2
In other words, the number of steps in the euclidean algorithm is bounded by a
linear function in the number of digits of a.
Proof. First we make an important obervation. Apply the first step of the
euclidean algorithm to obtain q, r ∈ Z≥0 such that a = bq + r with 0 ≤ r < b.
Then we assert that r < a/2. Indeed, if b > a/2 then necessarily q = 1 and
r = a − b < a − a/2 = a/2. If b ≤ a/2 we have automatically r < b ≤ a/2.
This observation can also be applied to any step ri−1 = qi ri +ri+1 in the euclidean
algorithm. Hence ri+1 < ri−1 /2 for every i. By induction we now find r1 <
a/2, r3 < a/4, r5 < a/8, . . . and in general rl < a/2(l+1)/2 for every odd l. When
l is even we observe that l − 1 is odd, so rl < rl−1 < a/2l/2 . So we find for all
l ∈ N that rl < a/2l/2 . In particular, rk < a/2k/2 . Using the lower bound rk ≥ 1
our theorem now follows.
2
In the exercises√we will show that the better, and optimal, estimate k < log(a)/ log(η)
with η = (1 + 5)/2 is possible.
1.4
Fundamental theorem of arithmetic
Definition 1.4.1 A prime number is a natural number larger than 1, which has
only 1 and itself as positive divisor.
Usually the following theorem is taken for granted since it is basically taught at
elementary school. However, its proof requires some work and is an application
of Corollary 1.2.4(iv).
Theorem 1.4.2 (Fundamental theorem of arithmetic) Any integer larger
than 1 can be written uniquely, up to ordering of factors, as the product of prime
numbers.
F.Beukers, Elementary Number Theory
1.4. FUNDAMENTAL THEOREM OF ARITHMETIC
11
Proof. First we show that any n ∈ N>1 can be written as a product of primes.
We do this by induction on n. For n = 2 it is obvious. Suppose n > 2 and
suppose we proved our assertion for all numbers below n. If n is prime we are
done. Suppose n is not prime. Then n = n1 n2 , where 1 < n1 , n2 < n. By our
induction hypothesis n1 and n2 can be written as a product of primes. Thus n is
a product of primes.
We now prove our theorem. Let n be the smallest number having two different
prime factorisations. Write
n = p1 p2 · · · pr = q 1 q 2 · · · q s ,
where pi , qj are all primes. Notice that p1 |q1 · · · qs . Hence according to Corollary 1.2.4(iv), p1 |qt for some t. Since qt is prime we have p1 = qt . Dividing out
the common prime p1 = qt on both factorisations we conclude that n/p1 has also
two different factorisations, contradicting the minimality of n.
2
The proof of the above theorem relies on Corollary 1.2.4 of Theorem 1.2.3. The
latter theorem relies on the fact that Z is a euclidean domain. In Theorem 13.2.3
we have given the analogue of Theorem 1.2.3 for general euclidean domains and in
principle it is possible to prove a unique prime factorisation property for arbitrary
commutative euclidean domains. We have not done this here but instead refer to
standard books on algebra.
Definition 1.4.3 Let a1 , . . . , an ∈ Z be non-zero. The lowest common multiple
of a1 , . . . , an is the smallest positive common multiple of a1 , . . . , an .
Notation: lcm(a1 , . . . , an ) or [a1 , . . . , an ].
The following lemma is a straightforward application of Theorem 1.4.2,
Lemma 1.4.4 Let a, b ∈ N be non-zero. Write
a = pk11 pk22 · · · pkr r ,
mr
1 m2
b = pm
1 p 2 · · · pr ,
where p1 , . . . , pr are distinct primes and 0 ≤ ki , mj (i, j = 1, . . . r). Then,
min(k1 ,m1 )
gcd(a, b) = p1
r ,mr )
· · · pmin(k
r
and
max(k1 ,m1 )
lcm(a, b) = p1
r ,mr )
· · · pmax(k
.
r
Proof. Excercise.
Another fact which is usually taken for granted is that there are infinitely many
primes. However, since it does not occur in our Axioms, we must give a proof.
Theorem 1.4.5 (Euclid) There exist infinitely many primes.
Suppose that there exist only finitely many primes p1 , . . . , pn . Consider the number N = p1 · · · pn + 1. Let P be a prime divisor of N . Then P = pi for some i
and pi |(p1 · · · pn + 1) implies that pi |1. This is a contradiction, hence there exist
infinitely many primes.
2
F.Beukers, Elementary Number Theory
12
CHAPTER 1. INTEGERS AND THE EUCLIDEAN ALGORITHM
1.5
Exercises
Exercise 1.5.1 Prove, ∀n ∈ N :
induction on n)
13 + 23 + · · · + n3 = (1 + 2 + · · · + n)2 (use
Exercise 1.5.2 Prove, 2n − 1 is prime ⇒ n is prime.
(The converse is not true, as shown by 89|211 − 1.)
Exercise 1.5.3 Prove, 2n + 1 is prime ⇒ n = 2k , k ≥ 0.
(Fermat thought that the converse is also true. However, Euler disproved this by
5
showing that 641|22 + 1.)
Exercise 1.5.4 (**) Prove, n4 + 4n is prime ⇒ n = 1.
Exercise 1.5.5 Prove that there exist infinitely many primes p of the form p ≡
−1(mod 4). Hint: use a variant of Euclid’s proof.
Exercise 1.5.6 Prove that there exist infinitely many primes p such that p + 2
is not a prime.
Exercise 1.5.7 (*)Let x, y ∈ N. Prove that x + y 2 and y + x2 cannot be both a
square.
Exercise 1.5.8 Prove that n2 ≡ 1(mod 8) for any odd n ∈ N.
Exercise 1.5.9 Determine d = (4655, 12075) and determine x, y ∈ Z such that
d = 4655x + 12075y.
Exercise 1.5.10 Let a, b be relatively prime positive integers and c ∈ Z and
consider the equation
ax + by = c
in the unknowns x, y ∈ Z.
1. Show that there exists a solution.
2. Let x0 , y0 be a solution. Show that the full solution set is given by
x = x0 + bt,
y = y0 − at
where t runs through Z.
3. Solve the equation 23x + 13y = 3 in x, y ∈ Z completely
Exercise 1.5.11 Let a, b be any positive integers and c ∈ Z and consider the
equation
ax + by = c
in the unknowns x, y ∈ Z.
F.Beukers, Elementary Number Theory
1.5. EXERCISES
13
1. Show that the equation has a solution if and only if gcd(a, b) divides c.
2. Describe the full solution set of the equation.
3. Solve the equation 105x + 121y = 3 in x, y ∈ Z completely.
Exercise 1.5.12 Let a, b ∈ N and suppose a > b. Choose r, q ∈ Z such that
a = bq + r and 0 ≤ r < b.
a) Show that r < a/2.
Consider now the euclidean algoritm with the notations from the course notes.
Let rk be the last non-zero remainder.
√
b) Prove that k < log a/ log 2.
Using the following steps we find an even better estimate for k,
√
c) Prove by induction on n = k, k − 1, . . . , 1 that rn ≥ ( 12 (1 + 5))k−n .
√
d) Prove that k < log a/ log( 12 (1 + 5)).
Exercise 1.5.13 To any pair a, b ∈ N there exist q, r ∈ Z such that a = qb + r
and |r| ≤ b/2. Based on this observation we can consider an alternative euclidean
algorithm.
a) Carry out this algorithm for a = 12075 and b = 4655.
b) Suppose a > b and let rk be the last non-zero remainder. Show that k <
log a/ log 2.
Exercise 1.5.14 Let a, m, n ∈ N and a ≥ 2. Prove that (am − 1, an − 1) =
a(m,n) − 1.
Exercise 1.5.15 (*) Let α be a positive rational number. Choose the smallest
integer N0 so that α1 = α − 1/N0 ≥ 0. Next choose the smallest integer N1 > N0
such that α2 = α1 − 1/N1 ≥ 0. Then choose N2 , etcetera. Show that there exists
an index k such that αk = 0. (Hint: First consider the case when α0 < 1).
Conclude that every positive rational number α can be written in the form
1
1
1
α=
+
+ ··· +
N0 N1
Nk−1
where N0 , N1 , . . . , Nk−1 are distinct positive integers.
Exercise 1.5.16 In how many zeros does the number 123! end?
Exercise 1.5.17 Let p be prime and a ∈ N. Define vp (a) = max{k ∈ Z≥0 | pk |a}.
a) Prove that
[ ] [ ] [ ]
n
n
n
vp (n!) =
+ 2 + 3 + · · · ∀n.
p
p
p
b) Let n ∈ N and write n in base p, i.e. write n = n0 + n1 p + n2 p2 + · · · + nk pk
with 0 ≤ ni < p for all i. Let q be the number of ni such that ni ≥ p/2. Prove
that
(( ))
2n
vp
≥ q.
n
F.Beukers, Elementary Number Theory
14
CHAPTER 1. INTEGERS AND THE EUCLIDEAN ALGORITHM
Exercise 1.5.18 Let b, c ∈ N and suppose (b, c) = 1. Prove,
a) For all a ∈ Z: (a, bc) = (a, b)(a, c).
b) For all x, y ∈ Z: (bx + cy, bc) = (c, x)(b, y).
Exercise 1.5.19 (**) Let a, b, c ∈ Z not all zero such that gcd(a, b, c) = 1. Prove
that there exists k ∈ Z such that gcd(a − kc, b) = 1.
Exercise 1.5.20 Prove or disprove: There are infinitely many primes p such
that p + 2 and p + 4 are also prime.
F.Beukers, Elementary Number Theory
Chapter 2
Arithmetic functions
2.1
Definitions, examples
In number theory we very often encounter functions which assume certain values
on N. Well-known examples are,
i. The unit function e defined by e(1) = 1 and e(n) = 0 for all n > 1.
ii. The identity function E defined by E(n) = 1 for all n ∈ N.
iii. The power functions Ik defined by Ik (n) = nk for all n ∈ N. In particular,
E = I0 .
iv. The number of prime divisors of n, denoted by Ω(n).
v. The number of distinct prime divisors of n, denoted by ω(n).
vi. The divisor sums σl defined by
σl (n) =
∑
dl .
d|n
In particular we write σ = σ1 , the sum of divisor and d = σ0 , the number
of divisors.
vii. The Euler ϕ-function or totient function
ϕ(n) = #{d ∈ N|gcd(d, n) = 1 and d ≤ n}.
viii. Ramanujan’s τ -function τ (n) defined by
∞
∑
n
τ (n)x = x
n=1
∞
∏
k=1
15
(1 − xk )24 .
16
CHAPTER 2. ARITHMETIC FUNCTIONS
ix. The ”sums of squares” function rd (n) given by the number of solutions
x1 , . . . , xd to n = x21 + · · · + x2d .
In general,
Definition 2.1.1 An arithmetic function is a function f : N → C.
Of course this is a very broad concept. Many arithmetic functions which occur
naturally have interesting additional properties. One of them is the multiplicative
property.
Definition 2.1.2 Let f be an arithmetic function with f (1) = 1. Then f is
called multiplicative if f (mn) = f (m)f (n) for all m, n with (m, n) = 1 and
strongly multiplicative if f (mn) = f (m)f (n) for all m, n.
It is trivial to see that examples e, E, Il , 2Ω are strongly multiplicative and that
2ω is multiplicative. In this chapter we will show that σl and ϕ are multiplicative.
The multiplicative property of Ramanujan’s τ is a deep fact based on properties
of so-called modular forms. It was first proved by Mordell in 1917. As an aside we
also mention the remarkable congruence τ (n) ≡ σ11 (n)(mod 691) for all n ∈ N.
The multiplicative property of r2 (n)/4 will be proved in the chapter on sums of
squares.
Theorem 2.1.3
i. σl (n) is a multiplicative function.
ii. Let n = pk11 · · · pkr r . Then
σl (n) =
∏ pl(ki +1) − 1
i
pl − 1
i
.
Proof. Part i. The proof is based on the fact that if d|mn and (m, n) = 1 then
d can be written uniquely in the form d = d1 d2 where d1 |m, d2 |n. In particular
d1 = (m, d), d2 = (n, d). We have
∑
∑
(d1 d2 )l
dl =
σl (mn) =
d1 |m,d2 |n
d|mn

= 
∑


∑
dl1  
dl2 
d1 |m
d2 |n
= σl (m)σl (n)
Part ii. It suffices to show that σl (pk ) = (pl(k+1) − 1)/(pl − 1) for any prime power
pk . The statement then follows from the multiplicative property of σl . Note that,
σl (pk ) = 1 + pl + p2l + · · · + pkl =
pl(k+1) − 1
.
pl − 1
2
A very ancient problem is that of perfect numbers.
F.Beukers, Elementary Number Theory
2.1. DEFINITIONS, EXAMPLES
17
Definition 2.1.4 A perfect number is a number n ∈ N which is equal to the sum
of its divisors less than n. Stated alternatively, n is perfect if σ(n) = 2n.
Examples of perfect numbers are 6, 28, 496, 8128, 33550336, ... . It is not known
whether there are infinitely many. It is also not known if there exist odd perfect
numbers. If they do, they must be at least 10300 . There is an internet search for
the first odd perfect number (if it exists), see /www.oddperfect.org.
For even perfect numbers there exists a characterisation given by Euclid and
Euler.
Theorem 2.1.5 Let n be even. Then n is perfect if and only if it has the form
n = 2k−1 (2k − 1) with 2k − 1 prime.
Proof. Suppose n = 2p−1 (2p − 1) with 2p − 1 prime. Then it is straightforward
to check that σ(n) = 2n.
Suppose that n is perfect. Write n = 2k−1 m, where m is odd and k ≥ 2. Then,
σ(n) = σ(2k−1 m) = σ(2k−1 )σ(m) = (2k − 1)σ(m).
On the other hand, n is perfect, so σ(n) = 2n, which implies that 2k m = (2k −
1)σ(m). Hence
m
σ(m) = m + k
.
2 −1
Since σ(m) is integral, 2k − 1 must divide m. Since k ≥ 2 we see that m and
m/(2k − 1) are distinct divisors of m. Moreover, they must be the only divisors
since their sum is already σ(m). This implies that m is prime and m/(2k −1) = 1,
i.e m = 2k − 1 is prime.
2
Numbers of the form 2m − 1 are called Mersenne numbers . If 2m − 1 is prime
we call it a Mersenne prime. It is an easy excercise to show that 2m − 1 prime
⇒ m is prime. Presumably there exist infinitely many Mersenne primes but this
is not proved yet. The known values of m for which 2m − 1 is prime, are
n=
2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203,
2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701,
23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433,
1257787, 1398269, 2976221, 3021377, 6972593, 13466917,
20996011, 24036583, 25964951, 30402457, 32582657,
37156667, 42643801, 42643801, 43112609
At the moment (August 2008) 243112609 − 1 is the largest known prime. For the
latest news on search activities see: www.mersenne.org.
An equally classical subject is that of amicable numbers that is, pairs of numbers
m, n such that n is the sum of all divisors of m less than m and vice versa. In
F.Beukers, Elementary Number Theory
18
CHAPTER 2. ARITHMETIC FUNCTIONS
other words, m+n = σ(n) and n+m = σ(m). The pair 220, 284 was known to the
ancient Greeks. Euler discovered some 60 pairs (for example 11498355, 12024045)
and later computer searches yielded several thousands of new pairs, some of which
are extremely large.
2.2
Convolution, Möbius inversion
Definition 2.2.1 Let f and g be two arithmetic functions. Their convolution
product f ∗ g is defined by
∑
(f ∗ g)(n) =
f (d)g(n/d).
d|n
It is an easy excercise to verify that the convolution product is commutative and
associative. Moreover, f = e ∗ f for any f . Hence arithmetic functions form a
semigroup under convolution.
Theorem 2.2.2 The convolution product of two multiplicative functions is again
multiplicative.
Proof. Let f, g be two multiplicative functions. We have trivially that (f ∗
g)(1) = f (1)g(1). For any m, n ∈ N with (m, n) = 1 we have
( mn )
∑
f (d)g
(f ∗ g)(mn) =
d
d|mn
(
)
∑∑
mn
f (d1 d2 )g
=
d1 d2
d1 |m d2 |n



(
(
)
)
∑
m  ∑
n 
f (d1 )g
= 
f (d2 )g
d1
d2
d1 |m
d2 |n
= (f ∗ g)(m)(f ∗ g)(n).
2
Notice that for example σl = E ∗ Il . The multiplicative property of σl follows
directly from the mulitiplicativity of E and Il . We now introduce an important
multiplicative function.
Definition 2.2.3 The Möbius function µ(n) is defined by µ(1) = 1, µ(n) = 0 if
n is divisible by a square > 1 and µ(p1 · · · pt ) = (−1)t for any product of distinct
primes p1 , . . . , pt .
Notice that µ is a multiplicative function. Its importance lies in the following
theorem.
F.Beukers, Elementary Number Theory
2.2. CONVOLUTION, MÖBIUS INVERSION
19
Theorem 2.2.4 (Möbius inversion) Let f be an arithmetic function and let
F be defined by
∑
F (n) =
f (d).
d|n
Then, for any n ∈ N,
f (n) =
∑
F (d)µ
d|n
(n)
d
.
Proof. More cryptically we have F = E ∗ f and we must prove that f = µ ∗ F .
It suffices to show that e = E ∗ µ since this implies µ ∗ F = µ ∗ E ∗ f = e ∗ f = f .
The function E ∗ µ is again multiplicative, hence it suffices to compute E ∗ µ at
prime powers pk and show that it is zero there. Observe,
(E ∗ µ)(pk ) =
∑
µ(d) = µ(1) + µ(p) + · · · + µ(pk )
d|pk
= 1 − 1 + 0 + · · · + 0 = 0.
2
Theorem 2.2.5 Let ϕ be the Euler ϕ-function. Then,
i.
n=
∑
ϕ(d),
∀n ≥ 1.
d|n
ii. ϕ is multiplicative.
iii.
ϕ(n) = n
∏(
p|n
1
1−
p
)
.
Proof. Part i). Fix n ∈ N. Let d|n and let Vd be the set of all m ≤ n such
that (m, n) = d. After dividing everything by d we see ∑
that |Vd | = ϕ(n/d).
∑ Since
{1, . . . , n} = ∪d|n Vd is a disjoint union we find that n = d|n ϕ(n/d) = d|n ϕ(d),
as asserted.
Part ii). We have seen in part i) that I1 = E ∗ ϕ. Hence, by Möbius inversion,
ϕ = µ ∗ I1 . Multiplicativity of ϕ automatically follows from the multiplicativity
of µ and I1 .
Part iii). Because of the multiplicativity of ϕ it suffices to show that ϕ(pk ) =
pk (1 − 1/p). This follows from ϕ(pk ) = (I1 ∗ µ)(pk ) = pk − pk−1 = pk (1 − p1 ). 2
F.Beukers, Elementary Number Theory
20
2.3
CHAPTER 2. ARITHMETIC FUNCTIONS
Exercises
Exercise 2.3.1 Prove that the sum of the inverses of all divisors of a perfect
number is two.
Exercise 2.3.2 Prove that an odd perfect number must contain at least three
distinct primes.
(It is known that there must be at least six distinct primes).
Exercise 2.3.3 Describe the multiplicative functions which arise from the following convolution products (the symbol 2ω stands for the multiplicative function
2ω(n) ).
Ik ∗ Ik
µ ∗ I1
µ∗µ
µ ∗ 2ω
2ω ∗ 2ω
µ ∗ ϕ.
Exercise 2.3.4 Prove that there exist infinitely many n such that µ(n) + µ(n +
1) = 0.
Exercise 2.3.5 Prove that there exist infinitely many n such that µ(n) + µ(n +
1) = −1.
Exercise 2.3.6 Let F be the multiplicative function such that F (n) = 1 if n
is a square, and 0 otherwise. Determine a multiplicative function f such that
f ∗ E = F.
Exercise 2.3.7 Let f be a multiplicative function. Prove that there exists a
multiplicative function g such that f ∗ g = e (hence the multiplicative functions
form a group with respect to the convolution product).
Exercise 2.3.8 Prove that for every n ∈ N
2

∑
∑
σ0 (k) .
σ0 (k)3 = 
k|n
k|n
Exercise 2.3.9 (**) Show that there exist infinitely many n such that ϕ(n) is a
square.
F.Beukers, Elementary Number Theory
Chapter 3
Residue classes
3.1
Basic properties
Definition 3.1.1 Let m ∈ N. Two integers a and b are called congruent modulo
m if m|a − b.
Notation: a ≡ b(mod m)
Definition 3.1.2 The residue class a mod m is defined to be the set {b ∈ Z | b ≡
a(mod m)}. The set of residue classes modulo m is denoted by Z/mZ.
The following properties are more or less straightforward,
Remark 3.1.3 Let a, b, c, d ∈ Z and m, n ∈ N.
i. a ≡ a + rm(mod m) ∀r ∈ Z.
ii. There are m residue classes modulo m.
iii. a ≡ b(mod m) and n|m ⇒ a ≡ b(mod n).
iv. a ≡ b(mod m) and c ≡ d(mod m) ⇒ a + c ≡ b + d(mod m), ac ≡
bd(mod m).
v. Let P (x) ∈ Z[x]. Then a ≡ b(mod m) ⇒ P (a) ≡ P (b)(mod m).
vi. The residue classes modulo m form a commutative ring with 1-element.
Its additive group is cyclic and generated by 1 mod m. If m is prime then
Z/mZ is a finite field. If m is composite then Z/mZ has divisors of zero.
Definition 3.1.4 Let m ∈ N. A residue class a mod m is called invertible if
∃ x mod m such that ax ≡ 1(mod m). The set of invertible residue classes is
denoted by (Z/mZ)∗ .
21
22
CHAPTER 3. RESIDUE CLASSES
Notice that (Z/mZ)∗ is nothing but the unit group of Z/mZ. In particular the
inverse x mod m of a mod m is uniquely determined modulo m.
Theorem 3.1.5 Let a ∈ Z en m ∈ N. Then,
a mod m is invertble ⇐⇒ (a, m) = 1.
Proof. The residue class a mod m is invertible ⇐⇒ ∃x : ax ≡ 1(mod m) ⇐⇒
∃x, y : ax + my = 1 ⇐⇒ (a, m) = 1.
2
Notice that the proof of the latter theorem also shows that we can compute
the inverse of a residue class via the euclidean algorithm. For example, solve
331x ≡ 15(mod 782). The euclidean algorithm shows that the g.c.d. of 331 and
782 is 1 and 1 = −189 · 331 + 80 · 782. Hence −189 mod 782 is the inverse of
331 mod 782. To solve our question, multiply on both sides with −189 to obtain
x ≡ 293(mod 782).
3.2
Chinese remainder theorem
Theorem 3.2.1 Let m, n ∈ Z and (m, n) = 1. Then the natural map ψ :
Z/mnZ → Z/mZ × Z/nZ given by ψ : a mod mn 7→ (a mod m, a mod n) yields
an isomorphism of the rings Z/mnZ and Z/mZ × Z/nZ.
Proof. It is clear that ψ is a ringhomomorphism. Furthermore ψ is injective,
for if we assume ψ(a) = ψ(b) this means that a and b are equal modulo m and
n. So m and n both divide a − b and since (m, n) = 1, mn divides a − b, i.e
a ≡ b mod mn. Also, Z/mnZ and Z/mZ × Z/nZ have the same cardinality, mn.
Since an injective map between finite sets of the same cardinality is automatically
bijective, our theorem follows.
2
Via almost the same proof we obtain the following generalisation.
Theorem 3.2.2 Let m1 , . . . , mr ∈ Z and (mi , mj ) = 1 ∀i ̸= j. Let m =
m1 · · · mr . Then the map
ψ : Z/mZ → (Z/m1 Z) × · · · × (Z/mr Z)
given by
ψ : a mod m 7→ a mod m1 , . . . , a mod mr
yields a ring isomorphism.
A direct consequence of Theorem 3.2.2 is as follows
F.Beukers, Elementary Number Theory
3.2. CHINESE REMAINDER THEOREM
23
Corollary 3.2.3 (Chinese remainder theorem) Let notations be the same
as in Theorem 3.2.2. To any r-tuple of residue classes c1 mod m1 , . . . , cr mod mr
there exists precisely one residue class a mod m such that a ≡ ci (mod mi ) (i =
1, . . . , r).
Let us give a slightly different reformulation and another proof.
Theorem 3.2.4 (Chinese remainder theorem) Let m1 , . . . , mr ∈ Z and (mi , mj ) =
1 ∀i ̸= j. Let m = m1 · · · mr . Then the system of equations
x ≡ cj (mod mj ),
1≤j≤r
has exactly one residue class modulo m as solution set.
Proof. Define Mj = m/mj for j = 1, . . . , r. Then clearly (Mj , mj ) = 1, ∀j.
Choose for each j a number nj such that Mj nj ≡ 1(mod mj ). Let
x0 =
r
∑
cj nj Mj .
j=1
Using mi |Mj if i ̸= j and ni Mi ≡ 1(mod mi ) we can observe that x0 ≡ ci (mod mi )
for any i. Clearly, any number congruent to x0 modulo m is also a solution of
our system.
Conversely, let x1 be a solution of our system. Then x1 ≡ x0 mod mj and hence
mj |(x1 − x0 ) for all j. Since (mi , mj ) = 1 for all i ̸= j, this implies m|(x1 − x0 )
and hence x1 ≡ x0 mod m, as desired.
2
Notice that the latter proof of the Chinese remainder theorem gives us an algorithm to construct solutions. However, in most cases it is better to use straightforward calculation as in the following example.
Example. Solve the system of congruence equations
x ≡ 1(mod 5),
x ≡ 2(mod 6),
x ≡ 3(mod 7).
First of all notice that 5, 6 and 7 are pairwise relatively prime and so, according to
the chinese remainder theorem there is a unique residue class modulo 5·6·7 = 210
as solution. However, in the following derivation this will follow automatically.
The first equation tells us that x = 1 + 5y for some y ∈ Z. Substitute this
into the second equation, 1 + 5y ≡ 2(mod 6). Solution of this equation yields
y ≡ 5(mod 6). Hence y must be of the form y = 5 + 6z for some z ∈ Z.
For x this implies that x = 26 + 30z. Substitute this into the third equation,
26 + 30z ≡ 3(mod 7). Solution yields z ≡ 6(mod 7). Hence z = 6 + 7u, u ∈ Z
which implies x = 206 + 210u. So the residueclass 206 mod 210 is the solution to
our problem. Notice by the way that 206 ≡ −4(mod 210). If we would have been
clever we would have seen the solution x = −4 and the solution to our problem
without calculation.
F.Beukers, Elementary Number Theory
24
CHAPTER 3. RESIDUE CLASSES
3.3
Invertible residue classes
In this section we shall give a description of the group (Z/mZ)∗ and a number
of properties. With the notations of Theorem 3.2.2 we know that Z/mZ ≃
Z/m1 Z × · · · × Z/mr Z. This automatically implies the following corollary.
Corollary 3.3.1 Let notations be the same as in Theorem 3.2.2. Then,
(Z/mZ)∗ ≃ (Z/m1 Z)∗ × · · · × (Z/mr Z)∗ .
Definition 3.3.2 Let m ∈ N. The number of natural numbers ≤ m relatively
prime with m is denoted by ϕ(m), Euler’s totient function.
Notice that if m ≥ 2 then ϕ(m) is precisely the cardinality of (Z/mZ)∗ . We recall
the following theorem which was already proved in the chapter on arithmetic
functions. However, the multiplicative property follows also directly from the
chinese remainder theorem. So we can give a separate proof here.
Theorem 3.3.3 Let m, n ∈ N.
a) If (m, n) = 1 then ϕ(mn) = ϕ(m)ϕ(n).
∏
b) ϕ(n) = n (1 − p1 ), where the product is over all primes p dividing n.
∑
c)
d|n ϕ(d) = n.
Proof. Part a) follows directly from Corollary 3.3.1 which implies that (Z/mnZ)∗ ≃
(Z/mZ)∗ × (Z/nZ)∗ . Hence |(Z/mmZ)∗ | ≃ |(Z/mZ)∗ | × |(Z/nZ)∗ | and thus
ϕ(mn) = ϕ(m)ϕ(n).
Part b) First note that ϕ(pk ) = pk − pk−1 for any prime p and any k ∈ N. It is
simply the number of integers in [1, pk ] minus the number of multiples of p in the
same interval. Then, for any n = pk11 · · · pkr r we find, using property (a),
ϕ(n) =
r
∏
ϕ(pki i )
i=1
=
r
∏
(pki i − pki i −1 ) = n
i=1
∏
(1 −
i
Part c) Notice that
n = #{k|1 ≤ k ≤ n}
∑
=
#{k|1 ≤ k ≤ n, (k, n) = d}
d|n
=
∑
#{l|1 ≤ l ≤ n/d, (l, n/d) = 1}
d|n
=
∑
ϕ(n/d) =
d|n
F.Beukers, Elementary Number Theory
∑
d|n
ϕ(d)
1
).
pi
3.3. INVERTIBLE RESIDUE CLASSES
25
Theorem 3.3.4 (Euler) Let a ∈ Z and m ∈ N. Suppose (a, m) = 1. Then
aϕ(m) ≡ 1(mod m).
Proof. Since (a, m) = 1, the class a mod m is an element of (Z/mZ)∗ , a finite
abelian group. Since ϕ(m) = #(Z/mZ)∗ , our statement follows from Lemma 13.1.1 2
When m is a prime, which we denote by p, we have ϕ(p) = p − 1 and thus the
following corollary.
Corollary 3.3.5 (Fermat’s little theorem) Let a ∈ Z and let p be a prime.
Suppose p ̸ |a. Then
ap−1 ≡ 1(mod p).
Corollary 3.3.5 tells us that 2p−1 ≡ 1(mod p) or equivalently, 2p ≡ 2(mod p),
for any odd prime p. Almost nothing is known about the primes for which
2p ≡ 2(mod p2 ). Examples are given by p = 1093 and 3511. No other such
primes are known below 3 × 109 . It is not even known whether there exist
finitely or infinitely many of them. And to add an even more remarkable token
of our ignorance, neither do we know if there infinitely many primes p with
2p ̸≡ 2(mod p2 ).
Theorem 3.3.6 (Wilson,1770) If p is a prime then (p − 1)! ≡ −1(mod p).
Proof. Notice that this theorem is just a special case of Lemma 13.1.12, when
we take R = (Z/pZ)∗ .
2
Theorem 3.3.7 (Gauss) Let p be a prime. Then the group (Z/pZ)∗ is cyclic.
Proof. Since p is prime, (Z/pZ)∗ is the unit group of the field Z/pZ. From
Lemma 13.1.11 our statement follows.
2
Definition 3.3.8 Let m ∈ N. An integer g such that g mod m generates the
group (Z/mZ)∗ is called a primitive root modulo m.
As an illustration consider (Z/17Z)∗ and the powers of 3 modulo 17,
31 ≡ 3
32 ≡ 9 33 ≡ 10 34 ≡ 13 35 ≡ 5 36 ≡ 15
37 ≡ 11 38 ≡ 16 39 ≡ 14 310 ≡ 8 311 ≡ 7 312 ≡ 4
313 ≡ 12 314 ≡ 2 315 ≡ 6 316 ≡ 1
Observe that the set {31 , . . . , 316 } equals the set {1, 2, . . . , 16} modulo 17. So 3
is a prinitive root modulo 17. Notice also that 24 ≡ 16 ≡ −1(mod 17). Hence
28 ≡ (−1)2 ≡ 1(mod 17) and 2 is not a primitive root modulo 17.
F.Beukers, Elementary Number Theory
26
CHAPTER 3. RESIDUE CLASSES
Not all values of m allow a primitive root. For example, the order of each elements
in (Z/24Z)∗ is 1 or 2, whereas this group contains 8 elements. To get a good
impression of (Z/mZ)∗ one should take the prime decomposition m = pk11 · · · pkr r
of m and notice that according to Corollary 3.3.1
(Z/mZ)∗ ≃ (Z/pk11 Z)∗ × · · · × (Z/pkr r Z)∗ .
The structure for the groups (Z/pk Z)∗ with p prime and k ∈ N is then given in
the following two theorems.
Theorem 3.3.9 Let p be an odd prime and k ∈ N. Then (Z/pk Z)∗ is a cyclic
group.
Proof. For k = 1 the theorem is just Theorem 3.3.7. So let us assume k > 1.
Let g be a primitive root modulo p and let ord(g) be its order in (Z/pk Z)∗ . Since
g ord(g) ≡ 1(mod pk ) we have g ord(g) ≡ 1(mod p). Moreover, g is a primitive root
modulo p and thus (p − 1)|ord(g). So h = g ord(g)/(p−1) has order p − 1 in (Z/pk Z)∗ .
k−1
From Lemma 3.3.11(i) with r = k it follows that (1 + p)p ≡ 1(mod pk ) and the
k−2
same lemma with r = k − 1 implies (1 + p)p
≡ 1 + pk−1 ̸≡ 1(mod pk ). Hence
ord(1 + p) = pk−1 . Lemma 13.1.6 now implies that (p + 1)h has order (p − 1)pk−1
and so (Z/pk Z)∗ is cyclic.
2
Theorem 3.3.10 Let k ∈ Z≥3 . Any element of (Z/2k Z)∗ can be written uniquely
in the form (−1)m 5t mod 2k with m ∈ {0, 1}, 0 ≤ t < 2k−2 .
Note that this theorem implies that (Z/2k Z)∗ is isomorphic to the product of a
cyclic group of order 2 and a cyclic group of order 2k−2 when k ≥ 3. Of course
(Z/4Z)∗ and (Z/2Z)∗ are cyclic.
Proof. From Lemma 3.3.11(ii) with r = k we find 52
≡ 1(mod 2k ) and with
2k−3
k−1
k
r = k − 1 we find 5
≡1+2
̸≡ 1(mod 2 ). So, ord(5) = 2k−2 . Notice that
all elements 5t , 0 ≤ t < 2k−2 are distinct and ≡ 1(mod 4). Hence the remaining
elements of (Z/2k Z)∗ are given by −5t , 0 ≤ t < 2k−2 .
2
k−2
Lemma 3.3.11 Let p be an odd prime and r ∈ N.
r−1
i. (1 + p)p
r−2
ii. 52
≡ 1 + pr (mod pr+1 ).
≡ 1 + 2r (mod 2r+1 ) for all r ≥ 2.
Proof. i. We use induction on r. For r = 1 our statement is trivial. Let r > 1
and assume we proved
r−2
(1 + p)p
≡ 1 + pr−1 (mod pr ).
F.Beukers, Elementary Number Theory
3.4. PERIODIC DECIMAL EXPANSIONS
r−2
In other words, (1 + p)p
on both sides,
pr−1
(1 + p)
= 1 + Apr−1 with A ≡ 1(mod p). Take the p-th power
= 1+
p ( )
∑
p
t=1
≡ 1 + pAp
Because p is odd we have
27
(p)
2
r−1
(1 + p)p
t
r−1
(Apr−1 )t
( )
p
+
(Apr−1 )2 (mod pr+1 ).
2
≡ 0(mod p) and we are left with
≡ 1 + Apr ≡ 1 + pr (mod pr+1 )
as asserted.
ii. Use induction on r. For r = 2 our statement is trivial. Let r > 2 and assume
we proved
r−3
52 ≡ 1 + 2r−1 (mod 2r ).
r−3
In other words, 52
52
r−2
= 1 + A2r−1 with A odd. Take squares on both sides,
= 1 + A2r + A2 22r−2 ≡ 1 + 2r (mod 2r+1 ).
The latter congruence follows because A is odd and 2r − 2 ≥ r + 1 if r > 2. This
concludes the induction step.
2
3.4
Periodic decimal expansions
A very nice application of the properties of residue classes is that in writing
decimal expansions of rational numbers. For example, when we write out the
decimal expansion of 1/7 we very soon find that
1
= 0.142857142857142857142857 . . . ,
7
in short hand notation
1
= 0./14285/7.
7
Using Euler’s Theorem 3.3.4 this is easy to see. Note that 106 − 1 is divisible by
7 and 106 − 1 = 7 · 142857. Then,
142857
1
= 6
= 142857 · 10−6 + 142857 · 10−12 + · · ·
7
10 − 1
The second equality is obtained by expanding 1/(106 − 1) = 10−6 /(1 − 10−6 ) in
a geometrical series.
F.Beukers, Elementary Number Theory
28
CHAPTER 3. RESIDUE CLASSES
Definition 3.4.1 A decimal expansion is called periodic if it is periodic from
a certain decimal onward, and purely periodic if it is periodic starting from the
decimal point.
For any periodic decimal expansion there is a minimal period and it is clear that
the minimal period divides any other period.
Theorem 3.4.2 Let α be a real number then,
i. The decimal expansion of α is periodic ⇔ α ∈ Q.
ii. The decimal expansion of α is purely periodic ⇔ α ∈ Q and the denominator
of α is relatively prime with 10.
Suppose α ∈ Q has a denominator of the form 2k 5l q, where (q, 10) = 1. Then
the minimal period length of the decimal expansion of α equals the order of 10 in
(Z/qZ)∗ .
Proof. Suppose that the decimal expansion of α is periodic. By multiplication
with a sufficiently large power 10k we can see to it that 10k α is purely periodic.
Suppose the minimal period length is l. Then there exist integers A, N such that
0 ≤ N < 10l − 1 and 10k α = A + N · 10−l + N · 10−2l + · · ·. The latter sum equals
A + N/(10l − 1), hence α is rational. In case the decimal expansion is purely
periodic we can take k = 0 and we have α = A + N/(10l − 1). Thus it is clear
that the denominator q of α is relatively prime with 10.
Suppose conversely that α ∈ Q. After multiplication by a sufficiently high power
10k we can see to it that the denominator q of 10k α is relatively prime with 10.
Then we can write 10k α = A + a/q, where A, a ∈ Z and 0 ≤ a < q. Let r be the
order of 10 in (Z/qZ)∗ and write N = a · (10r − 1)/q. Then 0 ≤ N < 10r − 1 and
10k α = A +
N
.
−1
10r
After expansion into a geometrical series we see that 10k α has a purely periodic
decimal expansion and that α has a periodic expansion. In particular, when the
denominator of α is relatively prime with 10 we can take k = 0 and the expansion
of α is purely periodic in this case.
To prove the last part of the theorem we make two observations. Let again r
be the order of 10 in (Z/qZ)∗ and let l be the minimal period of the decimal
expansion. From the first part of the above proof it follows that q divides 10l − 1.
In other words, 10l ≡ 1(mod q) and hence r|l. From the second part of the proof
it follows that r is a period of the decimal expansion, hence l|r. Thus it follows
that r = l.
2
The above theorem is stated only for numbers written in base 10. It should be
clear by now how to formulate the theorem for numbers written in any base.
F.Beukers, Elementary Number Theory
3.5. EXERCISES
3.5
29
Exercises
Exercise 3.5.1 Let n ∈ N be composite and n > 4. Prove that (n − 1)! ≡
0(mod n).
Exercise 3.5.2 Prove that a natural number is modulo 9 equal to the sum of its
digits.
Exercise 3.5.3 Prove that any palindromic number of even length is divisible by
11. (A palindromic number is a number that looks the same whether you read
from right to left or from left to right).
Exercise 3.5.4 Determine the inverse of the following three residue classes:
5(mod 7), 11(mod 71), 86(mod 183).
Exercise 3.5.5 A flea jumps back and forth between the numerals on the face of
a clock. It is only able to make jumps of length 7. At a certain moment the flea
is located at the numeral I. What is the smallest number of jumps required to
reach XII?
Exercise 3.5.6 Determine all x, y, z ∈ Z such that
a) x ≡ 3(mod 4)
b) 3y ≡ 9(mod 12)
c) z ≡ 1(mod 12)
x ≡ 5(mod 21)
x ≡ 7(mod 25)
4y ≡ 5(mod 35)
z ≡ 4(mod 21)
6y ≡ 2(mod 11)
z ≡ 18(mod 35).
Exercise 3.5.7 Determine all integral multiples of 7 which, after division by
2,3,4,5 and 6 yield the remainders 1,2,3,4 and 5 respectively.
Exercise 3.5.8 Prove that there exist a million consecutive positive integers,
each of which is divisible by a square larger than 1.
Exercise 3.5.9 Notice that 3762 ≡ 376(mod 103 ), 906252 ≡ 90625(mod 105 ).
a) Let k ∈ N. How many solutions a ∈ Z with 1 < a < 10k does a2 ≡ a(mod 10k )
have?
b) Solve the equation in a) for k = 12.
Exercise 3.5.10 For which n ∈ N does the number nn end with 3 in its decimal
expansion?
Exercise 3.5.11 (XXI st International Mathematics Olympiad 1979). Let N, M ∈
N be such that (N, M ) = 1 and
1 1 1
1
1
1
N
= 1 − + − + ··· +
−
+
.
M
2 3 4
1317 1318 1319
Show that 1979 divides N . Generalise this.
F.Beukers, Elementary Number Theory
30
CHAPTER 3. RESIDUE CLASSES
Exercise 3.5.12 Write the explicit isomorphism between Z/10Z and Z/2Z ×
Z/5Z.
Exercise 3.5.13 a) Prove that 13|(42
9
b) Prove that 37/|x9 + 4 ∀x ∈ N.
2n+1
− 3) ∀n ∈ N.
Exercise 3.5.14 Suppose m = pk11 · · · pkr r with pi distinct primes and ki ≥ 1 ∀i.
Put
λ(m) = lcm[pk11 −1 (p1 − 1), . . . , pkr r −1 (pr − 1)].
Show that aλ(n) ≡ 1(mod n) for all a such that (a, n) = 1.
Exercise 3.5.15 Show that n13 ≡ n(mod 2730) for all n ∈ Z.
Exercise 3.5.16 For which odd primes p is (2p−1 − 1)/p a square ?
Exercise 3.5.17 Let p be a prime such that p ≡ 1(mod 4). Prove, using Wilson’s
theorem, that ((p − 1)/2)!)2 ≡ −1(mod p).
Exercise 3.5.18 Is the following statement true: every positive integer, written
out decimally, is either prime or can be made into a prime by changing one digit.
Exercise 3.5.19 Find a prime divisor of 1111111111111 and of 111 . . . 111 (79
ones).
Exercise 3.5.20 Prove that for all a, n ∈ N with a ̸= 1 we have n|ϕ(an − 1).
Exercise 3.5.21 Prove that ϕ(n) → ∞ as n → ∞.
Exercise 3.5.22 (H.W.Lenstra jr.) Let N = 16842752.
a) Find an even n ∈ N such that ϕ(n) = N .
b) Prove that there is no odd number n ∈ N such that ϕ(n) = N .
c) Let 1 ≤ k ≤ 31. Construct an odd m ∈ N such that ϕ(m) = 2k .
Exercise 3.5.23 Prove that for all n ≥ 2,
n−1
∑
m=1
(m,n)=1
1
m = nϕ(n).
2
Exercise 3.5.24 Prove that
{
∑
0 if n ∈ N is even
(−1)n/d ϕ(d) =
−n if n ∈ N is odd
d|n
F.Beukers, Elementary Number Theory
3.5. EXERCISES
31
Exercise 3.5.25 a) Determine all solutions to ϕ(n) = 24.
b) Show that there are no solutions to ϕ(n) = 14.
Exercise 3.5.26 Given a ∈ N characterise all n ∈ N such that ϕ(n)/n = ϕ(a)/a.
Exercise 3.5.27 Determine all n ∈ N such that ϕ(n)|n.
Exercise 3.5.28 Compute the orders of 5(mod 7), 3(mod 46), 26(mod 17) in
the multiplicative groups mod m.
Exercise 3.5.29 Determine all primes such that thet period of the decimal expansion of 1/p is
a) at most 6
b) precisely 7
c) precisely 8.
Exercise 3.5.30 Let p be an odd prime and q a prime divisor of 2p − 1. Prove
that q = 2mp + 1 for some m ∈ N.
Exercise 3.5.31 Let n ∈ N, a ∈ Z and q an odd prime divisor of a2 + 1.
a) Prove that q = m · 2n+1 + 1 for some m ∈ N.
b) Prove that for any n ∈ N there exist infinitely many primes of the form p ≡
1(mod 2n ).
n
Exercise 3.5.32 Let q be an odd prime and n ∈ N. Show that any prime divisor
p of nq−1 + · · · + n + 1 satisfies either p ≡ 1(mod q) or p = q. Using this fact
show that there exist infinitely many primes which are 1 modulo q.
Exercise 3.5.33 Determine all primitive roots modulo 11. Determine all primitive roots modulo 121.
Exercise 3.5.34 a) Determine all primitive roots modulo 13, 14 and 15 respectively.
b) Solve the following equations: x5 ≡ 7(mod 13), x5 ≡ 11(mod 14), x5 ≡
2(mod 15).
Exercise 3.5.35 Suppose that (Z/mZ)∗ is cyclic. Prove that m = 2, 4, pk or 2pk ,
where p is an odd prime and k ∈ N. (Hint: use Exercise 3.5.14)
Exercise 3.5.36 Suppose that p and q = 2p + 1 are odd primes. Prove that
p−1
(−1) 2 2 is a primitive root modulo q.
Exercise 3.5.37 Show that 22 − 1 is divisible by at least t distinct primes.
t
F.Beukers, Elementary Number Theory
32
CHAPTER 3. RESIDUE CLASSES
Exercise 3.5.38 In the decimal expansion of 1/5681 the 99-th digit after the
decimal point is a 0. Prove this.
Exercise 3.5.39 What is the 840-th digit after the decimal point in the decimal
expansion of 1/30073?
Exercise 3.5.40 What is the 165-th digit after the decimal point in the decimal
expansion of 1/161?
Exercise 3.5.41 (Lehmer’s prime test). Let a, n ∈ N be such that
an−1 ≡ 1(mod n) and
a
n−1
p
̸≡ 1(mod n) ∀primes p|(n − 1).
Prove that n is prime.
Exercise 3.5.42 Use Lehmer’s test to prove that 3 · 28 + 1 is prime. (Hint:use
a = 11). Notice that Lehmer’s test is very suitable for numbers such as h · 2k + 1
when h is small.
F.Beukers, Elementary Number Theory
Chapter 4
Primality and factorisation
4.1
Prime tests and compositeness tests
It is well known that one can find the prime√factorisation of a numbers N by the
naive method of trying all numbers below N as divisor. It is also known that
it is extremely time-consuming to factor any number with more than 10 digits in
this way. As an example, try to factor N = 204906161249 in this way by hand!
When the number of digits of N exceeds 20 digits this naive method consumes
hours of computer time, even on a fast PC. During the last 20 years powerful
new methods have been developed to factorise large numbers more efficiently. A
recent success has been the factorisation of the Fermat number 2512 + 1, which
is now, and probably for a long time to be, the largest Fermat number of which
we know the complete prime factorisation. Nowadays (around 2012) numbers
of about 200 digits can be factored fairly routinely. Before one tries to factor a
number however, it makes sense to check first whether it is prime. Such primality
tests have been developed as well and using routine methods on a supercomputer
one can test primality of numbers up to 1000 digits. In this section we describe
a few simple tests which are in practice very efficient and can be programmed
easily on a PC.
Consider our number N again. We would like to know √
whether it is prime or
not. Again, the naive method of trying all numbers below N as divisor becomes
impossible if N contains more than 20 digits. The new idea is to use the following
contrapositive to Fermat’s little theorem.
Theorem 4.1.1 Let N ∈ N and let a ∈ Z be such that N does not divide a.
Then,
aN −1 ̸≡ 1(mod N ) =⇒ N is composite.
So, given N , pick any a not divisible by N and perform the above test. However,
before trying to apply this test, it is sensible to check whether (a, N ) = 1. If not,
N is of course composite and we have also found a divisor. In that case there is
33
34
CHAPTER 4. PRIMALITY AND FACTORISATION
no need to test N any further. As an example of the evaluation of aN −1 (mod N ),
take a = 2 and N = 204906161249. The computation of 2N −1 mod N may seem
formidable at first sight, but with a little thought it can be done very efficiently.
To do this write N − 1 in base 2,
N − 1 = 10111110110101010111110010001100000.
We perform a number of squaring operations and multiplications by 2 in such a
way, that we try to form the binary expansion of N − 1 in the exponent of 2 as
we go along,
21
210
2101
21011
2N −1
≡
≡
≡
≡
2(mod N )
22 ≡ 4(mod N )
2 · 42 ≡ 32(mod N )
2 · 322 ≡ 2048(mod N )
···
≡ 201135347146(mod N ).
The number of steps required by this method is precisely the number of binary
digits of N − 1 which is√proportional to log N . For large N this is extremely
small compared to the N steps required by the naive method. By the way,
we see that 2N −1 ̸≡ 1(mod N ), hence N is composite (it will turn out that
N = 369181 × 555029). So Theorem 4.1.1 can be considered as a compositeness
test. What to do if we had found 2N −1 ≡ 1(mod N ) instead? We cannot conclude
that N is prime. We have for example 2560 ≡ 1(mod 561), whereas 561 = 3·11·17.
But we can always choose other a and repeat the test. If aN −1 ≡ 1(mod N ) for
several a, we still cannot conclude that N is prime. It is becoming more likely,
but not 100% certain. In fact there exist N such that aN −1 ≡ 1(mod N ) for
all a with (a, N ) = 1. These are the so-called Carmichael numbers. Examples
are 561, 1729, 294409, . . .. There exist 2163 Carmichael numbers below 25 · 109 .
It was an exciting surprise when Granville, Pomerance and Red Alford showed
around 1991 that there exist infinitely many of them.
A criterion which gives better chances (but not certainty) in recognising primes
is based on the following refinement of Fermat’s little theorem.
Theorem 4.1.2 Let p be an odd number and a ∈ Z such that p ̸ |a. Write
p − 1 = 2k · m with m odd and k ≥ 0. Suppose p is prime. Then,
either am ≡ 1(mod p)
or ∃i such that a2 ·m ≡ −1(mod p) and 0 ≤ i ≤ k − 1.
i
Proof. Since p is prime we know that
k ·m
ap−1 ≡ a2
F.Beukers, Elementary Number Theory
≡ 1(mod p).
4.1. PRIME TESTS AND COMPOSITENESS TESTS
35
Suppose that am ̸≡ 1(mod p). Let r be the smallest non-negative integer such
r
r−1
that a2 ·m ≡ 1(mod p). Notice that 1 ≤ r ≤ k. Then a2 ·m ≡ ±1(mod p).
r−1
r−1
By the minimality of r we cannot have a2 ·m ≡ 1(mod p). Hence a2 ·m ≡
−1(mod p) and since 0 ≤ r − 1 ≤ k − 1 our assertion follows.
2
The contrapositive statement can be formulated as follows.
Theorem 4.1.3 (Rabin test) Let N ∈ N be odd and a ∈ Z such that N /
|a.
Write N − 1 = 2k · m with k ≥ 0 and m odd. If
am ̸≡ 1(mod N )
and
∀0≤i≤k−1 : a2 ·m ̸≡ −1(mod N )
i
(4.1)
then N is composite.
If a satisfies property (4.1) we shall call a a witness to the compositeness of
N . Unlike the converse to Fermat’s little theorem the Rabin test allows no
Carmichael-like numbers N . This is garantueed by the following theorem
Theorem 4.1.4 Let N ∈ N be odd and composite. Among the integers between
1 and N at least 75% is a witness to the compositeness of N .
Proof. It suffices to prove that at least 75% of the numbers in (Z/N Z)∗ is a
witness. We shall do this for all N ̸= 9. For N = 9 the theorem is directly verified
by hand. Let S be the union of the solution sets of the equations xm ≡ 1(mod N )
i
and x2 ·m ≡ −1(mod N )(i = 0, . . . , k − 1) respectively. It suffices to show that
|S| is at most ϕ(N )/4.
j
Let j be the largest number with 0 ≤ j ≤ k −1 such that x2 ·m ≡ −1(mod N ) has
a solution. Such a j exists since we have trivially (−1)m ≡ −1(mod N ). Notice
j
that S is contained in the set of solutions of x2 ·m ≡ ±1(mod N ). Now apply
Lemma 4.1.5 to see that there are at most ϕ(N )/4 such solutions.
2
Lemma 4.1.5 Let N be an odd composite positive integer, not equal to 9. Let
M |(N − 1)/2. Suppose that xM ≡ −1(mod N ) has a solution x0 . Then the
number of solutions to xM ≡ ±1(mod N ) in x ∈ (Z/N Z)∗ is less than or equal
to ϕ(N )/4.
Proof. Let N = pk11 · · · pkr r be the prime factorisation of N . Notice that the
number of solutions to xM ≡ 1(mod N ) equals the numbers of solutions to xM ≡
−1(mod N ), a bijection being given by x 7→ xx0 (mod N ). The total number of
solutions is equal to
∏
∏
2 (M, pki i −1 (pi − 1)) = 2 (M, pi − 1).
i
i
The equality follows from the fact that, since M |(N − 1), prime factors of N
cannot divide M . For every prime p dividing N we know that xM ≡ −1(mod p)
F.Beukers, Elementary Number Theory
36
CHAPTER 4. PRIMALITY AND FACTORISATION
has a solution, hence ord(x) divides 2M but not M . This implies (M, p − 1) =
ord(x)/2 ≤ (p − 1)/2.
Suppose that N has at least three distinct prime factors. Then,
∏
∏ pi − 1
1∏
ϕ(N )
2 (M, pi − 1) ≤ 2
≤
(pi − 1) ≤
.
2
4 i
4
i
i
Suppose that N has precisely two distinct prime factors, p and q say. First
suppose that N = pq. There exist e, f ∈ N such that p − 1 = 2e(M, p − 1) and
q − 1 = 2f (M, q − 1). Suppose e = f = 1. From e = 1 follows that (p − 1)/2
divides M which in its turn divides (N − 1)/2 = (pq − 1)/2. Hence (p − 1)/2
divides (q − 1)/2. Similarly it follows from f = 1 that (q − 1)/2 divides (p − 1)/2.
Hence p = q, which is impossible. So we must assume ef ≥ 2. But then,
2(M, p − 1)(M, q − 1) = 2
(p − 1)(q − 1)
(p − 1)(q − 1)
ϕ(N )
≤
=
.
4ef
4
4
Now suppose that N has two primes p, q and is divisible by p2 , say. Then,
2(M, p − 1)(M, q − 1) ≤
2(p − 1)(q − 1)
ϕ(N )
ϕ(N )
≤
≤
.
4
2p
4
Finally suppose that N is a prime power, pk , k ≥ 2. Then, since pk−1 ≥ 4,
2(M, p − 1) ≤ (p − 1) ≤
ϕ(N )
ϕ(N )
≤
.
k−1
p
4
2
In practice we choose a arbitrarily and apply Rabin’s test. Suppose that N
is composite. Let us assume that the witnesses to the compositeness of a are
distributed more or less randomly, which does not seem to be unreasonable.
Then, according to Theorem 4.1.3 the chance that N will not be recognised as
composite is less than 1/4. This means that after a 100, say, of such trials the
chance that N is not recognised as composite is less than (1/4)100 , very small
indeed. So, alternatively, if a number N is not recognised as composite after a
100 trials, the likelyhood that it is prime is practically 1!
Under the assumption of a certain conjecture Rabin’s test can be made into a
fullfledged primality test.
Theorem 4.1.6 Let N be a composite number. If the Generalised Riemann Hypothesis (GRH) is true, then there is witness to the compositeness of N which is
smaller than 2(log N )2 .
In other words, under the assumption of GRH, if N has no witness below 2(log N )2
then it is prime. This is known as the Rabin-Miller test and its run-time has the
pleasant property of being polynomial in the length of the input (i.e. polynomial
in log N ). Unfortunately, there is not much hope of proving the GRH yet.
A nice primality test is the following.
F.Beukers, Elementary Number Theory
4.1. PRIME TESTS AND COMPOSITENESS TESTS
37
Theorem 4.1.7 (Lehmer) Let N ∈ N. Suppose that we have a number a such
that
• aN −1 ≡ 1(mod N )
• a(N −1)/q ̸≡ 1(mod N ) for every prime q dividing N − 1.
Then N is prime.
Proof. Let k be the order of a(mod N ). From the first property it follows that k
divides N −1. Suppose that k equals N −1. Then we conclude that ϕ(N ) ≥ N −1
and this is only possible if N is prime. So we are done when k = N − 1.
Now suppose k < N − 1. Then there exists a prime divisor p of N − 1 such
that a(N −1)/p ≡ 1(mod N ). But this contradicts our second assumption on a. So
k < N − 1 cannot occur.
2
For numbers of a very special form there may exist very powerful primality tests
tailored to the special circumstances. We mention here the Fermat numbers
n
22 + 1 and the Mersenne numbers 2n − 1. The following theorem can be seen
as a very special case of Lehmer’s test where N − 1 has only prime factors 2. It
turns out that a = 3 has the desired properties.
Theorem 4.1.8 (Pepin, 1877) Let n ∈ N and Fn = 22 + 1. Then,
n
Fn is prime ⇔ 3
Fn −1
2
≡ −1(mod Fn ).
Theorem 4.1.9 (Lucas, 1876) Let Mn = 2n −1. Consider the sequence S1 , S2 , S3 , . . .
2
given by S1 = 4 and Sk = Sk−1
− 2 for all k ≥ 2. Then,
Mn is prime ⇔ Sn−1 ≡ 0(mod Mn ).
The proofs of these theorems are given in the chapter on quadratic reciprocity.
It turns out that F1 , F2 , F3 , F4 are prime. This led Fermat to believe that all
numbers Fn are prime. Ironically it turns out that Fn is composite for n =
5, 6, . . . , 22 and many other n. It is not even known whether Fn is prime for any
n > 4. Here is a list of factorisations of Fn for 5 ≤ n ≤ 9,
F5 = 641 · 6700417 (Euler, 1732)
F6 = 274177 · 67280421310721 (Landry, Le Lasseur, 1880)
F7 = 59649589127497217 ·
·5704689200685129054721 (Morrison, Brillhart, 1974)
F8 = 1238926361552897 ·
·(9346163971535797776916355819960689658405123754163
8188580280321) (Brent, Pollard, 1980)
F.Beukers, Elementary Number Theory
38
CHAPTER 4. PRIMALITY AND FACTORISATION
F9 = 2424833 ·
·7455602825647884208337395736200454918783366342657 ·
·(7416400626753080152478714190193747405994078109751
9023905821316144415759504705008092818711693940737)
(Lenstra, Manasse, 1990)
In 1644 The French monk Marin Mersenne stated that 2n − 1 is prime for the
values
n = 2, 3, 5, 7, 13, 17, 19, 31, 67, 127, 257
and composite for all other n < 257. It was clear that Mersenne had not tested
all these numbers. It was only in 1750 when Euler verified that 231 − 1 is prime.
It also turned out that Mersenne was wrong about n = 67, 257 and that he had
forgotten to add n = 61, 89, 107 to his list. The number 2127 − 1 was proven to be
prime in 1876 by Lucas and until 1952 this remained the largest known prime. For
further details and more on prime numbers see the Web page primes.utm.edu.
For the following values of n the number Mn is now (2013) known to be prime:
n=
2, 3, 5, 7, 13, 17, 19, 31, 61, 89, 107, 127, 521, 607, 1279, 2203,
2281, 3217, 4253, 4423, 9689, 9941, 11213, 19937, 21701,
23209, 44497, 86243, 110503, 132049, 216091, 756839, 859433,
1257787, 1398269, 2976221, 3021377, 6972593, 13466917,
20996011, 24036583, 25964951, 30402457, 32582657,
37156667, 42643801, 43112609, 57885161
At the moment (2013) 257885161 − 1 is the largest known prime.
For the latest news on search activities see: primes.utm.edu/mersenne/ or
www.mersenne.org.
4.2
A polynomial time primality test
In recent years, starting from the 1980’s, several powerful primality tests have
been invented. We mention the test of Adleman,Rumely,Lenstra and Cohen,
which uses Gauss sums (see the chapter on Gauss sums) and tests which use the
addition structure on elliptic curves and abelian varieties. These methods are
still used in practice.
Despite all these ingeneous developments it still was not clear whether there exists a primality test whose runtime is polynomial in the number of digits of the
number to be tested. This changed in July 2002. An Indian computer scientist,
M.Agrawal and two of his students, N.Kayal and N.Saxena, had discovered a
polynomial time primality test. This was a historical breakthrough in the theory of factorisation and primality proving. Another remarkable feature of their
F.Beukers, Elementary Number Theory
4.3. FACTORISATION METHODS
39
discovery was its elementary nature. The original ingredients were some modular arithmetic with polynomials and a deep theorem in analytic number theory.
However, through the efforts of H.W.Lenstra jr, the analytic number theory part
has been replaced by a property of prime numbers which will actually be proved
in these course notes.
To show that the algorithm is quite simple we give it here is pseudo-code.
Input: integer n > 1
1. If n = ab for a ∈ N and b > 1:
output Composite
2. Find the smallest r ∈ N such that ordr (n) > 4(log n)2 .
3. If 1 < gcd(a, n) < n for some a ≤ r:
output Composite
4. If n ≤ r, output Prime
√
5. For a = 1 to [2 ϕ(r) log n] do
If (X + a)n ̸= X n + a(mod X r − 1, n):
output Composite
6. output Prime
Here is a proof for the correctness of the algorithm. In Step 1. it is clear that if
n = ab with b > 1, then n is composite. Furthermore, in Step 3 it is clear that
if gcd(a, n) is not 1 or n, the number n is composite. For Step 5 we remark that
if n is prime, then (X + a)n ≡ X n + a(mod n) for all integers a. So, a fortiori,
(X + a)n ≡ X n + a(mod X r − 1, n) for all r and a. If this condition is violated
for some a, r, then of course n cannot be prime.
The main point is now to show that √
if (X + a)n ≡ X n + a(mod X r − 1, n) for our
chosen r and all a between 1 and [2 ϕ(r) log n], then n is prime.
4.3
Factorisation methods
The test of Rabin, described in the previous section, enables one to decide compositeness of a number N without knowing anything about the prime decomposition of N . In this section we make a few remarks about the problem of factoring
numbers. First of all, factoring a number is much harder than proving its compositeness. Another feature of factorisation algorithms is that they mostly give
a positive chance of success, but not certainty. Factorisation methods with 100%
of success are the naive method and the algorithm of Sherman-Lehman. However
they are both very slow with runtimes O(N 1/2 ) and O(N 1/3 ) respectively. One
therefore prefers the probabilistic methods with the filosophy that if you happen
F.Beukers, Elementary Number Theory
40
CHAPTER 4. PRIMALITY AND FACTORISATION
to find a factorisation of a large number you don’t complain about how it was
found. As an example of such a factorisation method we sketch the Pollard rho
algorithm. It works very well on a PC for numbers up to 25 digits.
The Pollard rho method. Choose b ∈ Z and b ̸= 0, −2. Compute the numbers
x0 , x1 , x2 , . . . by
x0 = 1 and xk+1 ≡ x2k + b(mod N ) for k ≥ 0
(4.2)
For each k check whether
N > gcd(xk − xl , N ) > 1
(4.3)
for some l < k. If condition (4.3) is fulfilled we have actually found a factor of N .
Let p be a prime divisor
√ of N . We assert that the probability to find p using this
algorithm with k < 2p√is larger than 1/2. Since a composite number N always
has a prime divisor < N , the expected run-time of our algorithm is therefore
O(N 1/4 ).
Here is a heuristic argument which supports our assertion. Consider the sequence
x0 , x1 , x2 , . . . modulo p. Then practical experience suggests that if b ̸= 0, −2, this
sequence behaves randomly modulo p. People are unable to prove this but it
seems like a good principle. Let q ∈ N. Being a random sequence, the probability
that the elements x0 , x1 , . . . , xq mod p are all distinct is
1
2
q
(1 − )(1 − ) · · · (1 − ).
p
p
p
Suppose that q >
√
2p. Then
q
∑
1
2
q
r
log(1 − )(1 − ) · · · (1 − ) =
log(1 − )
p
p
p
p
r=1
≃ −
q
∑
r
r=1
= −
p
1 q(q + 1)
< −1
2
p
and the probability
that x0 , . . . , xq mod p are distinct is less than e−1 < 1/2. So,
√
when q > 2p the probality that two elements in x0 , . . . , xq mod p are equal
is larger than√1/2. In other words, if we are moderately lucky we will find
0 ≤ l < k ≤ 2p such that xk ≡ xl (mod p). Note by the way, that the above
argument is an example of the so-called birthday paradox.
If we carry out the algorithm as described above, we would have to store the
elements xi of our sequence to verify (4.3). Moreover, testing all l < k with
F.Beukers, Elementary Number Theory
4.4. THE QUADRATIC SIEVE
41
respect to (N, xk − xl ) would considerably slow down the algorithm. In practice
there are two ways to overcome this problem.
The first is to compute
xi and x2i simultaneously at every step. Suppose there
√
exist l < k < 2p satisfying xk ≡ xl (mod p). By its very construction, the
sequence {xi mod p}∞
i=0 will be periodic from the index l onward with period
k − l. In particular, choose M ∈ N such that k > M (k − l) ≥ l. Then we have
by the periodicity, x2M (k−l) ≡ xM (k−l) (mod p) and
√ we find that p|(x2i − xi , N ) for
i = M (k − l). Notice that i = M (k − l) < k < 2p.
A second option is to compute the sequence of xi (mod N ) and whenever we hit
upon i being a power of 2 we save it as the number A. At every iteration we
check whether 1 < gcd(xi − A) < N . Suppose that from the index l onward the
sequence xi (mod p) becomes periodic with period k − l. Choose r minimal such
that 2r ≥ max l, k − l, then there exists i with i < 2r+1 such that xi − x2r ≡
xi − A ≡ 0(mod p).
It should also be noted that instead of xk+1 = x2k + b we could have chosen any
other recurrence which has a chance of producing random sequences modulo N .
Our choice is simply the simplest we could think of.
In many large factorisation programs one factors out small prime factors by native
trial division for primes up to 109 , say. As a second step one often uses the Pollard
rho method to find moderately small prime factors up to, say 1014 .
The factorisation algorithms which are the most powerful at the moment (2006)
are the quadratic sieve of Pomerance (1984, to be discussed in the next section),
the elliptic curve method of Lenstra (1984), the Number Field Sieve (1990) by
the Lenstra’s, Manasse and Pollard and variations on these methods. Typically,
these methods do not garantuee 100% success in factoring a number, and their
run-time analysis is again based on probability arguments. But most of the time
they are very successful.
4.4
The quadratic sieve
From time to time even Fermat was forced to factor large numbers during his
calculations. Of course he also stumbled upon the near impossibility of fcatoring
large numbers. However, Fermat did make a few observations which enabled his
to factor certain large numbers quickly. One such example is N = 8051. Fermat
noticed that this is the difference of two squares, 8051 = 902 − 72 and he got
8051 = (90 − 7)(90 + 7) = 83 · 97.
This idea can
√ be formalised as follows, which we call2 the method 2of Fermat.
Choose r = [ N ] and test if one of the numbers (r + 1) − N, (r + 2) − N, (r +
3)2 − N, . . . is a square. If this is the case, say (r + k)2 − N = m2 then we have
N = (r + k)2 − m2 = (r + k − m)(r + k + m) and thus a factorisation. The
difference between the factors is 2m. This means that if N = ab with a < b, then
F.Beukers, Elementary Number Theory
42
CHAPTER 4. PRIMALITY AND FACTORISATION
m = (b − a)/2. Hence
√
√
√
√
k = b − m − r = b − (b − a)/2 − [ N ] < (a + b)/2 − ab + 1 = 1 + ( b − a)2 /2.
From this we see that Fermat’s method cannot only work well if the difference
between a and b is small. If one is lucky this occurs and in this way Fermat
managed to factor numbers successfully. For example, the fcatorisation
2027651281 = 46061 · 44021
was found in only 12 steps. Unfortunately, for large differences Fermat’s method
fails completely. It may even be worse than naive factorisation.
In the history of number factoring there are many variations on Fermat’s method.
They all come down to the construction, in one way or another, of two integers
x and y such that x2 ≡ y 2 (mod N ). From the fact that
N = gcd(N, x2 − y 2 ) = gcd(N, x − y)gcd(N, x + y)
follows a possible non-trivial factorisation. The best method in this respect in
C.Pomerance’s quadratic sieve (1984).
√ This method starts in the same way as
Fermat’s method. We choose r = [ N ] + 1 and consider the numbers
(r + k)2 − N
k = 0, 1, 2, 3, . . .
We cannot assume that N is the product of two numbers of comparable size, so
we cannot expect to into a square very quickly. The idea however is that for
small k the numbers (r + k)2 − N are small relative to N . More precisely,
√
√
(r + k)2 − N < ( N + k)2 − N = 2k N + k 2 .
√
For small k the order of magnitude is k N . The philosophy is that smaller
numbers have a better chance of containing small prime factors.
We choose a bound B and search for numbers (r + k)2 − N all of whose prime
factors are ≤ B. To improve our chances we also
√ allow negative values of2k. Here
is an example. Take N = 123889. Then r = [ N ] + 1 = 352 and (r + k) − N =
15 + 704k + k 2 . We choose B = 13 and try all −20 ≤ k ≤ 20. We find the
following values of k for which k 2 + 704k + 15 consists of prime factors ≤ 13,
k
k 2 + 704k + 15
−19
−23 · 53 · 13
−24 · 36
−17
−9
−25 · 3 · 5 · 13
3·5
0
4
1
2 · 32 · 5
27 · 3 · 13
7
15
24 · 33 · 13
F.Beukers, Elementary Number Theory
4.4. THE QUADRATIC SIEVE
43
The idea is now to choose factorisation on the right hand side such their product is
a square. For example the products corresponding to k = −19, −9, 15 multiplied
yield a square. Explicitly,
(r − 19)2 ≡ −23 · 53 · 13(mod N )
(r − 9)2 ≡ −25 · 3 · 5 · 13(mod N )
(r + 15)2 ≡ 24 · 33 · 52 (mod N )
Multiplication of these congruences yields
(r − 19)2 (r − 9)2 (r + 15)2 ≡ (26 · 32 · 53 · 13)2 (mod N )
We have two different squares, equal modulo N . Let us see if this gives a factor
of N ,
gcd(N, (r − 19)(r − 9)(r + 15) − 26 · 32 · 53 · 13) = gcd(123889, 40982373) = 541
We were lucky and found N = 541 · 229.
From this example the principle of the method can be deduced. We first find
sufficiently many k so that (r + k)2 − N consists of prime factors ≤ B. By
sufficient we mean: at least two more values than the number of primes ≤ B,
preferably more. Linear algebra over Z/2Z tells us that we can choose from these
numbers a set whose product is a square. By taking this product modulo N we
find a relation of the form X 2 ≡ Y 2 (mod N ), from which we hope to deduce a
factorisation. The relationship with linear algebra can be seen from our example.
The factorisation table we gave before can be depicted schematically as follows.
k −1 2
−19 1 1
−17 1 0
−9
1 1
0
0 0
0 0
1
7
0 1
15
0 0
3
0
0
1
1
0
1
1
5
1
0
1
1
1
0
0
7
0
0
0
0
0
0
0
11
0
0
0
0
0
0
0
13
1
0
1
0
0
1
0
Notice that 7 and 11 do not occur in our factorisations. Finding factorsiations
whose product is a square comes down to finding rows in our table whose sum
is the zero-vector modulo 2. In other words, we must solve a system of linear
equations over Z/2Z.
It has not been explained yet why this method is fast. A careful analysis, using
deep heuristics from analytic number theory, shows that the expected runtime of
the algorithm is L(N )c where
)
(√
log(N ) log log(N )
L(N ) = exp
F.Beukers, Elementary Number Theory
44
CHAPTER 4. PRIMALITY AND FACTORISATION
and c = 2. The reader can verify that limN →∞ L(N )/N ϵ = 0 for every ϵ > 0. So
the runtime is not exponential, we call this a sub-exponential algorithm.
One of the bottle-necks of the algorithm is that for each (r + k)2 − N it must be
decided if it consists of prime factors ≤ B. It was Pomerance’s idea to replace this
by a sieving process, hence the name quadratic sieve. We work with a sequence
of number ak which we initially choose to be (r + k)2 − N . For every prime
p ≤ B we do the following. Solve (r + x)2 ≡ N (mod p) and let x1 , x2 be two
solutions with x1 ̸≡ x2 (mod p). For every k with with k ≡ x1 , x2 (mod p) we
replace ak by ak /p. Having done all this, we pick those k fro which ak = ±1. For
these k the number (r + k)2 − N consists only of primes ≤ B. Strictly speaking
we must also sieve for higher powers of p, but we leave this aside for the sake of
simplicity. The saving in runtime is tremendous, the runtime is now L(N ) instead
of L(N )2 . As a rule of thumb we choose B < L(N )b and |k| < L(N )a for suitable
0.1 < a, b < 1, depending on the implementation. A big advantage of the sieving
technique is that it can be distributed easily over different machines. To give an
idea, A.K.Lenstra’s factorisation of RSA-129 (129 decimals) was distributed over
1600 computers, mostly workstations and PC’s. The number of primes for which
there was sieved was 524339, about half a million. The lienar algebra part was
the solution of a system of linear equations modulo 2 in half a million variables.
This work must be done on a central computer.
4.5
Cryptosystems, zero-knowledge proofs
An important recent application of large primes is the cryptosystem of Rivest,
Shamir and Adleman (RSA-cryptosystem). This system consists of a public encryption key, by which anyone can encrypt messages or other information, and
a secret key which one uses to decrypt these messages again. The important
principle of the RSA-system is that one cannot deduce the secret key from the
public key. Let us give a short description and explanation.
Choose two prime numbers p, q and let N = pq. Let λ = ϕ(N ) = (p − 1)(q − 1).
Destroy the prime numbers. Notice that
aλ ≡ 1(mod N )
∀a with (a, N ) = 1.
Choose a k, l ∈ N such that kl ≡ 1(mod λ).
The public key consists of the numbers k and N , which may be advertised anywhere. The secret key consists of the numbers l and N , of which l is known only
to the owner. Write kl = 1 + mλ. The encryption goes as follows. Transform the
message into blocks of digits and consider them as numbers, of which we assume
that they are smaller than N . Let T be such a block. The encoded form of T
will be C determined by C ≡ T k (mod N ). The decryption simply consists of
determining C l mod N , because C l ≡ T kl ≡ T mλ+1 ≡ T (mod N ).
F.Beukers, Elementary Number Theory
4.5. CRYPTOSYSTEMS, ZERO-KNOWLEDGE PROOFS
45
Suppose we want to deduce the secret key l from the public key k. This could be
done as follows. Factorise N = pq, determine λ = (p − 1)(q − 1) and determine
l by solving kl ≡ 1(mod λ). Notice that in order to determine λ we need the
factorisation of N . No other methods are known. If however the secret primes p, q
contain about 100 digits, factorisation of N with the known methods is practically
impossible within a human or even universal lifetime. So the success of the RSAcryptosystem is based on the apparent inability of mathematicians to factor large
numbers.
Of course the above idea can be applied in the opposite direction. We then have
the possibility of electronic signatures. Let us sketch a very simplified application. An individual, say Peter, makes or buys his own secret/public key pair.
The public key is known to, say, a savings bank and the private key is securely
guarded by Peter. Suppose Peter wants to transfer $10,000.- from the bank to a
furniture shop electronically. He might write an electronic message saying ”My
name is Peter, please transfer $10,000.- to the account of furniture shop so and
so”. Unfortunately anyone could do this in Peter’s name with undesirable consequences for Peter. The solution is that Peter encrypts his message with his
private code and appends it to the message written in plain text. The bank then
decodes the scrambled message with the public key and observes that the result
corresponds with Peter’s plain text. Since no one else but Peter could have encrypted the message succesfully, the bank is convinced of the validity of Peter’s
message and carries out his order.
Other applications of modular arithmetic and the near impossibility to factor
large numbers are protocols by which one can prove that one possesses certain
information, a password for example, without revealing that information. Such
protocols are known as zero-knowledge proofs. . One such procedure, known as
identity proof, has been devised by Goldwasser,Micali and Rackoff in 1985.
Suppose Vincent calls Vera over the telephone and Vincent wants to convince Vera
that it is really him, which Vera must then verify. Of course Vincent can mention
a password, only known to him and Vera, to identify himself. However, there is
always the possibility of eavesdroppers. Another problem might be that Vincent
considers Vera too sloppy to confide his password to. Both these problems are
solved by using a zero-knowledge proof.
Just as above we let N be the product of two very large primes p and q. The secret
password of Vincent, only known to him, is a number a between 1 and N . The
name by which Vincent is known publicly is A, determined by A ≡ a2 (mod N )
and 1 < A < N . The identification protocol runs as follows. Vincent takes a
random number x and sends the square X ≡ x2 (mod N ) to Vera. Then Vera can
ask either of two questions,
i. send x
ii. send ax
F.Beukers, Elementary Number Theory
46
CHAPTER 4. PRIMALITY AND FACTORISATION
If Vincent is really Vincent he can of course do that and Vera can check that
x2 ≡ X(mod N ) in case i) and that (ax)2 ≡ AX(mod N ) in case ii).
What if an impostor tries to pass himself as Vincent? Before sending anything
he may try to guess Vera’s question. If he guesses i) then he can take any x send
X mod N and answer Vera’s question with x. If he guesses ii) he can take any
x, send A−1 X mod N and answer Vera’s question with x. In any case the chance
for the imposter to make the right guess is 1/2. However, if this question and
answer game is repeated a hundred times, say, the chance that the impostor is
not exposed as a fraud is (1/2)100 . This is small enough for Vera to be convinced
of Vincent’s identity if all questions are answered correctly. Notice that in the
process the value of a has not been revealed by Vincent.
Another possibility for the imposter is to infer Vincent’s password a from A. He
would have to solve x2 ≡ A(mod N ) and the only known way to do this is to
solve the equivalent system x2 ≡ A(mod p), x2 ≡ A(mod q). Taking square
roots modulo a prime is doable in practice (see chapter on quadratic reciprocity)
so it seems we are done. Unfortunately we need the factorisation of N again and
this turns out to be the bottleneck in solving x2 ≡ A(mod N ). Again the safety
of Vincent’s password relies on our inability to factor very large numbers.
F.Beukers, Elementary Number Theory
Chapter 5
Quadratic reciprocity
5.1
The Legendre symbol
In this chapter we shall consider quadratic equations in Z/mZ and study an
important criterion for the solubility of x2 ≡ a mod p, where p is an odd prime
(quadratic reciprocity).
Definition 5.1.1 Let p be an odd prime and a ∈ Z not divisible by p. Then a is
called a quadratic residue mod p if x2 ≡ a(mod p) has a solution and a quadratic
nonresidue modulo p if x2 ≡ a(mod p) has no solution.
Example. The quadratic residues modulo 13 read: 1,4,9,3,12,10 and the quadratic
nonresidues : 2,5,6,7,8,11.
Definition 5.1.2 Let p be an odd prime. The Legendre symbol is defined by

( )  1
if a is quadratic residue mod p
a
−1 if a is quadratic nonresidue mod p
=

p
0
if p|a
Theorem 5.1.3 Let p be an odd prime and a, b ∈ Z. Then
a) There are exactly
residues mod p.
p−1
2
quadratic residues mod p and
b) (Euler)
( )
p−1
a
≡ a 2 (mod p).
p
c)
( )( ) ( )
a
b
ab
=
.
p
p
p
47
p−1
2
quadratic non-
48
CHAPTER 5. QUADRATIC RECIPROCITY
d)
( )
1
= 1,
p
(
−1
p
)
{
=
1
if p ≡ 1(mod 4)
−1 if p ≡ −1(mod 4)
( )2
Proof. Part a). Consider the residue classes 12 , 22 , . . . , p−1
mod p. Since a2 ≡
2
(−a)2 (mod p) these are all quadratic residues modulo p. They are also distinct,
from a2 ≡ b2 (mod p) would follow a ≡ ±b(mod p) and when 1 ≤ a, b ≤ p−1
2
this implies a = b. So there are exactly p−1
quadratic
residues
modulo
p.
The
2
remaining p − 1 − p−1
= p−1
residu classes are of course quadratic nonresidues .
2
2
Part b) Clear if a ≡ 0(mod p). So assume a ̸≡ 0(mod p). Since (a(p−1)/2 )2 ≡
ap−1 ≡ 1(mod p) by Fermat’s little theorem we see that a(p−1)/2 ≡ ±1(mod p).
Suppose that a is a quadratic residue , i.e. ∃x such that x2 ≡ a(mod p). Then
1 ≡ xp−1 ≡ (x2 )(p−1)/2 ≡ a(p−1)/2 (mod p), which proves half of our assertion.
Since we work in the field Z/pZ, the equation x(p−1)/2 ≡ 1(mod p) has at most
p−1
solutions. We know these solutions to be the p−1
quadratic residues . Hence
2
2
(p−1)/2
a
≡ −1(mod p) for any quadratic nonresidue a mod p.
Part c)
( )( )
( )
p−1 p−1
p−1
a
b
ab
≡ a 2 b 2 ≡ (ab) 2 ≡
(mod p).
p
p
p
Because
(
) ( ) Legendre
( ) symbols can only be 0, ±1 and p ≥ 3, the strict equality
a
b
= ab
follows.
p
p
p
( )
( )
≡
Part d) Of course p1 = 1 is trivial. From part b) follows that −1
p
(−1)
5.2
p−1
2
(mod p). Since p ≥ 3 strict equality follows.
2
Quadratic reciprocity
One might wonder for which prime numbers the numbers 3 and 5 ,say, are
quadratic residue . Euler, Legendre and Gauss have occupied themselves with
this question. For example it turns out that
( )
( )
3
3
= 1 if p ≡ ±1(mod 12)
= −1 if p ≡ ±5(mod 12)
p
p
and
( )
5
= 1 if p ≡ ±1(mod 5)
p
( )
5
= −1 if p ≡ ±2(mod 5).
p
Starting from such observations Euler conjectured the quadratic reciprocity law
(see Theorem 5.2.6). Legendre gave an incomplete proof of it and later Gauss
managed to give several complete proofs. In this chapter we give a proof which
is basically a version given by Eisenstein. In the chapter on Gauss sums we shall
give another proof.
F.Beukers, Elementary Number Theory
5.2. QUADRATIC RECIPROCITY
49
Definition 5.2.1 The residue classes 1, 2, . . . , p−1
mod p are called positive, the
2
p−1
residue classes −1, −2, . . . , − 2 mod p are called negative.
Theorem 5.2.2 (Gauss’ Lemma) Let p be an odd prime and a ∈ Z not divisible by p. Then
( )
a
= (−1)µ
p
where µ is the number of negative residue classes from a, 2a, . . . , p−1
a mod p.
2
Proof. Consider the p − 1 residue classes ±a, ±2a, . . . , ± p−1
a mod p. They
2
are non-zero and mutually distinct. So, the sets {±a, ±2a, . . . , ± p−1
a} and
2
p−1
p−1
{±1, ±2, . . . , ± 2 } are equal modulo p. Of each pair ±1, ±2, . . . , ± 2 mod p
exactly one element occurs in the sequence a, 2a, . . . , p−1
a mod p. Thus we find
2
that
p−1
p−1
a · 2a · · ·
a ≡ (−1)µ 1 · 2 · · ·
(mod p)
2
2
and hence
)
(
)
(
p−1
p−1
p−1
µ
!a 2 ≡ (−1)
!(mod p).
2
2
p−1
2
After division by ( p−1
)! we obtain
≡ (−1)µ mod p and after using Theo2
( ) a
rem 5.1.3(b) we conclude that ap = (−1)µ .
2
Theorem 5.2.3 Let p be an odd prime. Then
( ) {
2
1
if p ≡ ±1(mod 8)
=
−1 if p ≡ ±3(mod 8)
p
Proof. We apply Gauss’ lemma. To do so we must count µ, the number of
negative residue classes among 2, 4, . . . , p − 1 mod p. So,
p+1
≤ n ≤ p − 1}
2
p+1
p−1
≤ n≤
}
= #{n|
4
2
µ = #{n even|
Replace n by
p+1
2
− n to obtain
µ = #{n | 1 ≤ n ≤
[
]
p+1
=
4
p+1
}
4
This implies that µ is even if p ≡ ±1(mod 8) and µ is odd if p ≡ ±3(mod 8).
Gauss’ lemma now yields our assertion.
2
F.Beukers, Elementary Number Theory
50
CHAPTER 5. QUADRATIC RECIPROCITY
Remark 5.2.4 Notice that
( )
p2 −1
2
= (−1) 8 .
p
Another consequence of Gauss’ lemma is the following lemma which will be
needed in the proof of the quadratic reciprocity law.
Lemma 5.2.5 Let p be an odd prime and a ∈ Z odd and not divisible by p.
Define
p−1
]
2 [
∑
as
S(a, p) =
.
p
s=1
( )
a
= (−1)S(a,p) .
p
( )
Proof. According to Gauss’ lemma we have ap = (−1)µ where µ is the number
Then
a mod p. Let 1 ≤ s ≤ p−1
. If
of negative residue classes among a, 2a, . . . , p−1
2
2
p−1
sa
sa mod p is a positive residue class we write sa = [ p ]p + us with 1 ≤ us ≤ 2 . If
sa mod p is a negative residue class we write sa = [ sa
]p+p−us with 1 ≤ us ≤ p−1
.
p
2
p−1
A straightforward check shows that {u1 , u2 , . . . , u p−1 } = {1, 2, . . . , 2 }. Addition
2
of these equalities yields
]
2 [
∑
sa
p−1
p−1
2
∑
sa = p
s=1
s=1
p
p−1
+ µp +
2
∑
(±us ).
s=1
Take both sides modulo 2,
p−1
2
∑
p−1
s ≡ S(a, p) + µ +
s=1
2
∑
us (mod 2)
s=1
p−1
≡ S(a, p) + µ +
2
∑
s (mod 2).
s=1
The summations
on both sides cancel and we are left with S(a, p) ≡ µ(mod 2)
( )
a
2
hence p = (−1)µ = (−1)S(a,p) .
Theorem 5.2.6 (Quadratic reciprocity law) Let p, q be two odd prime numbers. Then
( )( )
p−1 q−1
p
q
= (−1) 2 2 .
q
p
( )
( )
Alternatively pq = pq unless p ≡ q ≡ −1(mod 4), in which case we have
( )
( )
p
q
=
−
.
q
p
F.Beukers, Elementary Number Theory
5.3. A GROUP THEORETIC PROOF
51
Proof. Let S(a, p) be as in Lemma 5.2.5. Then we assert
S(q, p) + S(p, q) =
p−1q−1
.
2
2
To see this, picture the rectangle [0, p/2] × [0, q/2] and the lattice points (m, n) ∈
N2 with 1 ≤ m ≤ p−1
, 1 ≤ n ≤ q−1
inside it. The diagonal connecting (0, 0) and
2
2
(p/2, q/2) does not pass through any of the lattice points. Notice that the number
of lattice points below the diagonal is precisely S(q, p) and above the diagonal
q−1
lattice points, hence our assertion follows.
S(p, q). In total there are p−1
2
2
We can now combine our assertion with Lemma 5.2.5 to obtain
( )( )
p−1 q−1
p
q
= (−1)S(p,q)+S(q,p) = (−1) 2 2 .
q
p
2
5.3
A group theoretic proof
It is known that Gauss gave six (more or less) different proofs of the quadratic
reciprocity law. Since then the number of proofs has increased dramatically to an
estimated 200. The proof we have given above is essentially due to Eisenstein. In
P.Bachmann, Die Lehre von der Kreistheilung, de Gruyter, Berlin, Leipzig, 1921
we find several classical proofs and we can find another 25 in O.Baumgart, Über
das quadratische Reciprocitätsgesetz, Teubner, Berlin, 1885. In the article ”On
the quadratic reciprocity law” in J.Australian Math. Soc. 51(1991), 423-425 by
G.Rousseau we find a proof of the reciprocity law which is surprisingly simple if
one is acquainted with elementary group theory. It turns out to be an application
of the chinese remainder theorem and we like to present it here.
Another proof of Theorem 5.2.6. Let notations be as in the theorem. We
work in the group G = ((Z/pZ)∗ × (Z/qZ)∗ )/U where U = {(1, 1), (−1, −1)}.
Clearly {(i, j) | i = 1, 2, . . . , p − 1; j = 1, 2, . . . , (q − 1)/2} is a full set of representatives of G. Their product π equals
π ≡ ((p − 1)!(q−1)/2 , ((q − 1)/2)!p−1 ).
Since ((q − 1)/2)!2 ≡ (−1)(q−1)/2 (q − 1)!(mod q) we get
π ≡ ((p − 1)!(q−1)/2 , (−1)
q−1 p−1
2
2
(q − 1)!(p−1)/2 ).
Another full set of representatives of G is given by {(k(mod p), k(mod q)) | k =
1, 2, . . . , (pq − 1)/2; (k, pq) = 1}. This is a consequence of (Z/pqZ)∗ ≃ (Z/pZ)∗ ×
(Z/qZ)∗ (chinese remainder theorem). The product of these elements modulo p
equals
(∏p−1
) ∏(p−1)/2 q−1
q−3
( 2 p + i)
i=1 i(p + i)(2p + i) · · · ( 2 p + i)
i=1
1 · q · 2q · · · p−1
q
2
F.Beukers, Elementary Number Theory
52
CHAPTER 5. QUADRATIC RECIPROCITY
which equals
(p − 1)!
(q−1)/2
/q
(p−1)/2
≡ (p − 1)!
(q−1)/2
( )
q
(mod p).
p
Similarly we compute the product modulo q and we obtain
( ))
(
( )
p
q
(p−1)/2
(q−1)/2
, (q − 1)!
.
π ≡ (p − 1)!
p
q
Comparison of the two expressions for π yields
(
) (( q ) ( p )) ( ( q ) ( p ))
p−1 q−1
1, (−1) 2 2 ≡
,
≡ 1,
p
q
p
q
2
and hence the reciprocity law.
5.4
Applications
Example 1. Is x2 ≡ 84(mod 97) solvable? Notice that
( ) ( )( )( ) ( )( ) ( )( )
84
4
3
7
97
97
1
−1
=
=
=
= 1 · −1 = −1.
97
97
97
97
3
7
3
7
Hence our congruence equation is not solvable.
Example 2. Is 3x2 + 4x + 5 ≡ 0 mod 76 solvable? According to the Chinese
remainder theorem this congruence is equivalent to
3x2 + 4x + 5 ≡ 0(mod 4)
3x2 + 4x + 5 ≡ 0(mod 19).
The first equation is equivalent to x2 ≡ 1(mod 4), which is solvable. Multiply the
second by 13 on both sides to obtain x(2 +14x
+8
)
( ≡) 0(mod 19). After splitting off
3
squares, (x + 7)2 ≡ 3(mod 19). Since 19
= − 19
= −1, the second congruence
3
equation, and hence the original one, is not solvable.
Example 3. Let p be an odd prime. Then,
( ) {
−3
1
if p ≡ 1(mod 3)
=
−1 if p ≡ −1(mod 3).
p
This follows from
( ) ( )( )
(p) (p)
p−1
p−1
−3
−1
3
=
.
=
= (−1) 2 · (−1) 2
p
p
p
3
3
Since 1 is a quadratic residue modulo 3 and −1 a quadratic nonresidue our
assertion follows.
F.Beukers, Elementary Number Theory
5.4. APPLICATIONS
53
Example 4. Let En be the integer whose digits in base 10 consist of n ones,
e.g. E13 = 1111111111111. These numbers are known as repunits. Alternatively
En = (10n − 1)/9. As an example
( 10 )
(we
)like
( 5 )to show
( 67here
) that
(2E
) 33 is divisible by
2
67. We easily verify that 67 = 67 67 = − 5 = − 5 = 1. Hence, by
Theorem 5.1.3(b), 1033 ≡ 1 mod 67 and hence 67|E33 .
Extensive calculations show that among the numbers En with n < 50000 only
E2 , E19 , E23 , E317 , E1031
are prime and E49081 is probably prime (H.Dubner, 1999). Here are some factorisations,
E3
E5
E7
E11
E13
E17
=
=
=
=
=
=
111 = 3 · 37
41 · 271
239 · 4649
21649 · 513239
53 · 79 · 265371653
2071723 · 5363222357
Theorem 5.4.1 Let p be a prime such that p ≡ −1(mod 4) and 2p + 1 prime.
Then 2p + 1 divides 2p − 1.
(
)
2
Proof. Note that 2p + 1 is a prime which is 7(mod 8). Hence, 2p+1
= 1.
(
)
2
Theorem 5.1.3(b) now implies 2p ≡ 2p+1
≡ 1(mod 2p + 1).
2
As a corollary we see that the Mersenne numbers
223 − 1, 283 − 1, 2131 − 1
are not prime. For p < 10000 there are 100 values for which Theorem 5.4.1
applies. (See also Theorem 5.4.3 below and the chapter on applications of residues
for Mersenne numbers). We also note that if p is prime, then any prime divisor q
of 2p − 1 has the form q = 2pk + 1. So, when looking for prime divisors of 2p − 1
it makes sense to start by trying 2p + 1.
Theorem 5.4.2 (Pépin, 1877) For any n ∈ N let Fn = 22 + 1. Then,
n
Fn is prime ⇐⇒ 3 2 (Fn −1) ≡ −1(mod Fn ).
1
Proof. ‘⇒’ Because Fn is an odd prime we have
( ) ( ) ( )
1
3
Fn
−1
(F
−1)
n
≡
≡
≡
≡ −1(mod Fn ).
32
Fn
3
3
F.Beukers, Elementary Number Theory
54
CHAPTER 5. QUADRATIC RECIPROCITY
The second to last congruence follows from 22 + 1 ≡ (−1)2 + 1 ≡ 1 + 1 ≡
−1(mod 3).
n
‘⇐’ Notice that Fn − 1 = 22 and 3Fn −1 ≡ 1(mod Fn ). Hence ord(3) divides
n
22 and equals 2r for some 0 ≤ r ≤ 2n . Suppose r < 2n then we would have
3(Fn −1)/2 ≡ 1(mod Fn ), contradicting our assumption. Hence r = 2n and ord(3) =
Fn − 1. In general, if we have a ∈ Z such that ord(a) in (Z/mZ)∗ is m − 1 then
m must be prime. In particular, Fn is prime.
2
n
n
The numbers Fn are known as the Fermat numbers, see the chapter on applications of congruences for more on the primality of Fn .
Theorem 5.4.3 (Lucas, Lehmer) For any n ∈ N let Mn = 2n − 1. Define
S1 , S2 , . . . by the recursion
S1 = 4
Sk+1 = Sk2 − 2,
∀k ≥ 0.
If n ≥ 3 and odd then,
Mn is prime ⇐⇒ Mn divides Sn−1 .
In√the proof of Lucas’ criterion we shall work in rings of the form Rm := {a +
b 3| a, b√∈ Z/mZ} where m ∈ N. More strictly speaking, Rm := Z[X]/(m, X 2 −
3) or Z[ 3]/(m). We state one result in Rm separately.
√
Let p be a prime larger than 3. Let a, b ∈ Z. Then, (a + b 3)p ≡
Lemma
( ) 5.4.4
√
a + p3 b 3(mod p).
Proof. Since p is a prime we have, using Fermat’s little Theorem 3.3.5,
√
√
(a + b 3)p ≡ ap + bp ( 3)p (mod p)
p−1 √
≡ a + b · 3 2 3(mod p).
Our lemma now follows if we use Euler’s Theorem 5.1.3(b).
Proof of Theorem 5.4.3. By induction on k it is not hard to show that
√ k−1
√ k−1
∀k ≥ 1.
Sk = (2 + 3)2 + (2 − 3)2
The condition Mn |Sn−1 can be rewritten as follow,
√ n−2
√ n−2
Sn−1 ≡ 0(mod Mn ) ⇐⇒ (2 + √3)2 + (2 − 3)2
≡ 0(mod Mn )
n−1
⇐⇒ (2 + √3)2 + 1 ≡ 0(mod Mn )
n−1
⇐⇒ (2 + 3)2
≡ −1 (mod Mn ).
F.Beukers, Elementary Number Theory
2
5.5. JACOBI SYMBOLS, COMPUTING SQUARE ROOTS
55
√ 2n−2
The second equivalence
follows
by
multiplication
with
(2
+
3)
and using the
√
√
fact that (2 + 3)(2 − 3) = 1. The proof of our theorem now comes down to
proving that
√ n−1
Mn is prime ⇐⇒ (2 + 3)2
≡ −1(mod Mn ).
√
The latter congruence can also be written as (2 + 3)(Mn +1)/2 ≡ −1(mod Mn )
and we can now note the similarity with Pepin’s test.
‘⇒’. First notice that, n being odd and ≥ 3, we have Mn ≡ 7(mod 24). Hence
if Mn is prime then 3 is a quadratic nonresidue mod Mn and 2 is a quadratic
residue mod Mn . Since Mn is assumed prime we have according to Lemma 5.4.4,
(
)
√
√ M
3 √
n
3 ≡ 1 − 3(mod Mn ).
(1 + 3) ≡ 1 +
Mn
Thus we find,
√
√
√
√
√ n
(1 + 3)2 ≡ (1 + 3)Mn (1 + 3) ≡ (1 − 3)(1 + 3) ≡ −2(mod Mn ). (5.1)
On the other hand,
√ n
√
√ n−1
√ n−1
n−1
n−1
(1+ 3)2 ≡ (1+ 3)2·2
≡ (4+2 3)2
≡ 22 (2+ 3)2 (mod Mn ). (5.2)
Since 2 is a quadratic residue modulo Mn , we have
n−1
22
≡ 2 · 2(Mn −1)/2 ≡ 2(mod Mn ).
(5.3)
Combination of (5.1), (5.2), (5.3) finally yields
√ n−1
(2 + 3)2
≡ −1(mod Mn )
as asserted.
‘⇐’. Let p be a prime divisor of Mn such that p ̸≡ ±1(mod 12). Since Mn ≡
7(mod 12) such a p exists. In particular we
that 3√
is a quadratic nonresidue
√have
p
mod p. Hence Lemma
5.4.4
yields
(2
+
3)
≡
2
−
3(mod p). After multi√
√ p+1
plication by 2 + 3 we find (2√+ 3)
≡ 1(mod p). On the other hand, √
by
2n−1
≡ −1(mod p). This implies that 2 + 3
assumption we have that (2 + 3)√
has order 2n in the unit group of Z[ 3]/(p). Hence 2n divides p+1. But p divides
Mn = 2n − 1. Thus we conclude that p = Mn and Mn is prime.
2
5.5
Jacobi symbols, computing square roots
( 111 )
To determine the Legendre symbol 137
say, we must first factor 111 before being
able to apply quadratic reciprocity. This( is all right)for small numbers like 111,
11111111111
? (197002597249 is prime)
but what to do if we want to compute 197002597249
F.Beukers, Elementary Number Theory
56
CHAPTER 5. QUADRATIC RECIPROCITY
or Legendre symbols with even larger numbers? In the chapter on applications
of congruences we pointed out that factorisation of large numbers is a major
computational problem. Luckily this does not mean that the computation of
Legendre symbols becomes difficult. The solution is to use the slightly more
general Jacobi symbol.
Definition 5.5.1 Let n ∈ N be odd and m ∈ Z such that (m,(n)) = 1. Let
n = p1 p2 · · · pr be the prime factorisation of n. The Jacobi symbol m
is defined
n
by
(m) (m) (m) (m)
···
=
n
p1
p2
pr
( )
where the symbols m
are the Legendre symbols.
pi
( )
Remark 5.5.2 Note that if m
= −1 then x2 ≡ m(mod n) is not solvable
n
2
simply
( m ) because x ≡ m(mod pi ) is not solvable for some i. 2On the other hand,
anything about the solubility of x ≡ m(mod n). For
if n = (1 we
) cannot say
−1
2
example, 21 = 1 but x ≡ −1(mod 21) is certainly not solvable.
However, we do have the following theorem.
Theorem 5.5.3 Let n, m be odd positive integers. Then,
i)
(
ii)
iii)
−1
n
)
= (−1)
n−1
2
( )
n2 −1
2
= (−1) 8
n
(m) ( n )
n
m
= (−1)
m−1 n−1
2
2
Proof. These statements can be proved by using the corresponding theorems for
the Legendre symbol and the observation that for any r-tuple of odd numbers
u1 , . . . , ur we have
ur − 1
u 1 · · · ur − 1
u1 − 1
+ ··· +
≡
(mod 2).
2
2
2
(5.4)
To be more precise, the sum on the left of (5.4) is modulo 2 equal to the number k
of ui which are −1(mod 4). If k is even, the product u1 · · · ur is 1(mod 4) and the
term on the right of (5.4) is also even. If k is odd, we have u1 · · · ur ≡ −1(mod 4),
hence the term on the right of (5.4) is also odd.
F.Beukers, Elementary Number Theory
5.5. JACOBI SYMBOLS, COMPUTING SQUARE ROOTS
57
Let n = p1 · · · pr be the prime factorisation of n. Then i) follows from
( ) ( ) ( )
p1 −1
pr −1
−1
−1
−1
=
···
= (−1) 2 +···+ 2
n
p1
pr
and
p1 − 1
pr − 1
p1 · · · pr − 1
n−1
+ ··· +
≡
≡
(mod 2).
2
2
2
2
Similarly, ii) follows from
( ) ( )
( )
p2
p2
2
2
2
r −1
1 −1
=
···
= (−1) 8 +···+ 8
n
p1
pr
and
p21 − 1
p2 − 1
(p1 · · · pr )2 − 1
n2 − 1
+ ··· + r
≡
≡
(mod 2).
8
8
8
8
Let m = q1 · · · qs be the prime factorisation of m. Then iii) follows from
(m) ( n ) ∏ ( q ) (p )
∑ pj −1 qi −1
i
j
=
= (−1) i,j 2 2
n
m
pj
qi
i,j
and
∑ p j − 1 q i − 1 ∑ q i − 1 ∑ pj − 1
m−1n−1
≡
≡
(mod 2).
2
2
2
2
2
2
i,j
i
j
2
( 11111111111 )
The computation of 197002597249 can now be done using a euclidean-like algorithm and Theorem 5.5.3. Notice that
197002597249 = 17 · 11111111111 + 8113708362
8113708362 = 2 · 4056854181
11111111111 = 2 · 4056854181 + 2997402749
...
...
Hence,
(
) (
)
11111111111
197002597249
=
197002597249
11111111111
(
) (
)(
)
8113708362
2
4056854181
=
=
11111111111
11111111111
( 11111111111 )
11111111111
=
= ···
4056854181
We keep repeating these steps of inversion and extraction of factors 2 until we
find the value of the Jacobi symbol to be 1. From this algorithm we see that
F.Beukers, Elementary Number Theory
58
CHAPTER 5. QUADRATIC RECIPROCITY
computation of Jacobi symbols, and hence Legendre symbols, is polynomial in
the length of the input.
Let)p be a prime and a ∈ Z. Suppose that by compution of the Legendre symbol
(
a
we know that x2 ≡ a(mod p) is solvable. How do we find a solution? The
p
following algorithm is very fast, provided we have a quadratic nonresidue mod
p at our disposal. Assuming the truth of the so-called Generalised Riemann
Hypothesis there exists a quadratic nonresidue between 1 and 2(log p)2 and the
algorithm we describe is then polynomial in the length of the input. Although we
do not know for sure that there exist small quadratic nonresidues , the Riemann
hypothesis is still unproved, experience seems to confirm their existence. So in
practice the algorithm described here is quite fast.
Solution of x2 ≡ a(mod p), Tonelli’s algorithm. Let p − 1 = 2s · m, with m
odd. Since (Z/pZ)∗ is a cyclic group the group G of elements whose order divides
2s is also cyclic. To find a generator of this group we must determine a quadratic
nonresidue modulo p. This can be done by just trying 2, 3, 5, 7, . . .. One very soon
hits upon a quadratic nonresidue . Call it x0 . We assert that b ≡ xm
0 (mod p) is
p−1
2s
m·2s
a generator of G. First of all note that b ≡ x0
≡ x0 ≡ 1(mod p), hence
s
the order of b divides 2s . On the other hand, if the (order
) of b is less than 2 ,
then b is an even power of the generator of G, hence pb = 1. This contradicts
( ) ( )m
b
= xp0
= (−1)m = −1. Hence b generates G.
p
Let x be a solution of x2 ≡ a(mod p). Raise both sides to the power (m + 1)/2,
x · xm ≡ a(m+1)/2 (mod p). Note that xm ∈ G. Hence, up to an element of G,
the solution x equals a(m+1)/2 (mod p). So we simply try a(m+1)/2 bj as a solution
for j = 0, 1, . . . , 2s−1 − 1. So, if 2s is small we find a solution very soon by just
trying j = 0, 1, . . . , 2s−1 − 1. If 2s is large we can use an additional trick, due to
D.Shanks, which is described below.
As an example we solve x2 ≡ 11111111111(mod 197002597249). Write p =
197002597249. It turns out that 7 is the least quadratic nonresidue mod p. We
have p − 1 = 27 · 1539082791. Let m = 1539082791. Then,
b ≡ 7m ≡ 59255134607(mod p)
r ≡ 11111111111(m+1)/2 ≡ 68821647300(mod p).
All we have to do now is try the numbers bj r(mod p) with j = 0, 1, 2, . . . , 63 as
solution of x2 ≡ 11111111111(mod p). It turns out that j = 37 does the job and
we find
574553913082 ≡ 11111111111(mod 197002597249).
If, in Tonelli’s algorithm the value of 2s is large, we can speed up the last part of
the algorithm as follows (Shanks),
Let G be a cyclic group of order 2s with generator b. Let g ∈ G. Then the output
e of the following algorithm is precisely the number such that g = be .
F.Beukers, Elementary Number Theory
5.6. CLASS NUMBERS
59
e := 0
loop:
Choose t ∈ Z≥0 minimal such that g 2 = 1.
t
If (t > 0) g := gb−2
s−t
, e := e + 2s−t , goto loop
If (t == 0) stop
By using this algorithm with g = x2 a−m−1 we can determine the value of 2j in
Tonelli’s algorithm quickly.
5.6
Class numbers
Take an odd prime p > 3 and consider the sum of quadratic residues
∑
a.
R=
1≤a≤p−1,a residue mod p
Let N be the analogous sum of quadratic non-residues. Notice that
∑
(p−1)/2
R≡
k=1
1
k 2 ≡ p(p − 1)(2p − 1) ≡ 0(mod p).
6
Also,
R+N ≡
p−1
∑
1
a ≡ p(p − 1) ≡ 0(mod p).
2
a=1
Hence both R and N are divisible by p. Let us make a table of (N − R)/p.
p
(N − R)/p
5 7
0 1
11
1
13
0
17
0
19
1
23
3
29
0
31
3
37
0
41
0
43
1
47
5
53
0
59
3
This table suggests that (N −R)/p is always ≥ 0 and 0 if and only if p ≡ 1(mod 4).
Suppose first that p ≡ 1(mod 4). If a is a residue modulo p, the same holds for
p − a. So the quadratic residues mod p come in pairs and there are (p − 1)/4
such pairs. Moreover, the sum of each pair is p, hence R = p(p − 1)/4. The
same argument shows that N = p(p − 1)/4. This confirms our expectation that
N − R = 0 if p ≡ 1(mod 4). Proving that (N − R)/p > 0 if p ≡ 3(mod 4)
is far more difficult however. A well-known proof by Dirichlet uses arguments
from complex function theory. Here is a table of values when p ≡ 3(mod 4) and
p < 200.
p
(N − R)/p
7 11
1 1
19
1
23
3
31
3
43
1
47
5
59
3
67
1
71
7
79
5
83
3
103
5
F.Beukers, Elementary Number Theory
60
CHAPTER 5. QUADRATIC RECIPROCITY
p
107
(N − R)/p 3
127
5
131
5
139
3
151
7
163
1
167
11
179
5
191
9
199
9
When
[ √ p ]≡ 3(mod 4) we call (N − R)/p the class-number of the ring O =
Z 1+ 2 −p . Notation: h(−p). These class numbers form the tip of an iceberg,
which is the field of quadratic forms, arithmetic in quadratic fields and Dirichlet L-series. For example, the interest of this class number lies in the fact that
h(−p) = 1 if and only if we have unique factorisation in irreducibles in O. It was
already suspected by Gauss that the biggest p for which h(−p) = 1 is p = 163.
This was only proved in the 1950’s by Heegner, Stark and later, in the 1960’s via
other methods, by A.Baker.
5.7
Exercises
Exercise 5.7.1 Determine the quadratic residues modulo 17 and 19.
Exercise 5.7.2 Fermat observed that if a, b ∈ N and (a, b) = 1 and p is an odd
prime divisor of a2 + b2 , then p ≡ 1(mod 4). Prove this.
Exercise 5.7.3 Let p be an odd prime and a ∈ Z not divisible by p. Prove by
induction on k,
x2 ≡ a(mod p) has a solution ⇒ x2 ≡ a(mod pk )has a solution.
Exercise 5.7.4 Let p be a prime and suppose p ≡ 1(mod 8).
a) Prove that x4 ≡ −1(mod p) has a solution.
b) Choose a solution x of x4 ≡
p) and compute the residue class (x +
( −1(mod
)
2
−1 2
x ) (mod p). Conclude that p = 1 if p ≡ 1(mod 8).
Exercise 5.7.5 Let p be a prime and suppose p ≡ 1(mod 3).
1. Prove that x2 + x + 1 ≡ 0(mod p) has a solution.
( )
2. Prove, using a), that −3
= 1 if p ≡ 1(mod 3).
p
3. Determine the discriminant of x2 + x + 1.
(
4. Prove,using considerations such as in a) and b) that
−1(mod 3).
Exercise 5.7.6 Show that
( 105 )
131
−3
p
)
= −1 if p ≡
= 1 and solve x2 ≡ 105(mod 131).
F.Beukers, Elementary Number Theory
5.7. EXERCISES
61
Exercise 5.7.7 Verify which of the following equations are solvable,
x2
x2
x2
x2
x2
9x2 + 12x + 15
8x2
≡
≡
≡
≡
≡
≡
≡
114(mod 127)
61(mod 93)
47(mod 101)
47(mod 143)
837(mod 2996)
0(mod 58)
2x + 3(mod 175).
Exercise 5.7.8 For which prime numbers is 5 a quadratic residue ? Same question for −3 and 3.
Exercise 5.7.9 Let a ∈ Z and p, q two odd (
primes
) (not) dividing a. Prove,
a
a) If a ≡ 1(mod 4) then: p ≡ q(mod |a|) ⇒ p = aq .
( ) ( )
b) If a ≡ −1(mod 4) then: p ≡ q(mod 4|a|) ⇒ ap = aq .
Exercise 5.7.10 Let p be a prime which is −1 modulo 4. Let a be a quadratic
residue modulo p. Prove that a solution of x2 ≡ a(mod p) is given by x = a(p+1)/4 .
Exercise 5.7.11 Let p be a prime which is 5 modulo 8. Let r be an element
of (Z/pZ)∗ of order 4 and let a be a quadratic residue modulo p. Prove that a
solution of x2 ≡ a(mod p) is given by either x = a(p+3)/8 or x = ra(p+3)/8 .
Exercise 5.7.12 Prove that there exist infinitely many primes p such that p ≡
1(mod 4). Prove that there exist infinitely many primes p such that p ≡ −1(mod 4).
Exercise 5.7.13 Let p be an odd prime. Let a be the smallest positive integer
√
such that a is a quadratic nonresidue modulo p. Show that a < 1 + p.
Exercise 5.7.14 a) Let p = 4k + 1 be prime. Prove,
√
√
p2 − 1
√
[ p] + [ 2p] + · · · + [ kp] =
.
12
b) Let p = 4k + 3 be prime. Prove,
√
√
(p − 1)(p − 2)
√
.
[ p] + [ 2p] + · · · + [ kp] ≤
12
√
(Hint: count the number of lattice points below the parabola y = px with x < p/4
and use in b) that N ≥ K.)
F.Beukers, Elementary Number Theory
Chapter 6
Dirichlet characters and Gauss
sums
6.1
Characters
In this chapter we study an important tool in number theory, namely characters
on (Z/mZ)∗ .
Definition 6.1.1 Let G be a finite abelian group. A homomorphism χ : G → C∗
is called a character of G.
Since any element g of a finite group has finite order, χ(g) must have finite order
in C∗ , hence be a root of unity. In particular, χ(e) = 1 for all χ. Moreover if G
is cyclic and g is a generator of G, then χ is determined by its value χ(g). Any
other a ∈ G is of the form a = g k and so we must have χ(a) = χ(g k ) = (χ(g))k .
Definition 6.1.2 The trivial character or principal character on a finite abelian
group G is the character given by χ(g) = 1 ∀g ∈ G. Notation: χ0 .
Suppose we have two characters χ1 and χ2 on G. Then one easily verifies that the
new function χ1 χ2 given by (χ1 χ2 )(g) = χ1 (g)χ2 (g) ∀g ∈ G is again a character
on G. In fact, the characters on G form a group with the trivial character as
identity element. We denote this group by Ĝ. Correspondingly we can speak
about the order of a character as being its order in Ĝ.
Lemma 6.1.3 Let G be a finite abelian group and χ a non-trivial character on
G. Then,
∑
χ(g) = 0.
g∈G
62
6.1. CHARACTERS
63
Proof. Since χ is non-trivial, there exists h ∈ G such that χ(h) ̸= 1. We now
use the fact that if g runs through G then so does gh,
∑
∑
∑
χ(g) =
χ(hg) = χ(h)
χ(g).
g∈G
g∈G
g∈G
Because χ(h) ̸= 1 our assertion follows.
2
To get a good grasp on the characters of an abelian group we consider the vectorspace V = C|G| consisting of |G|-tuples of complex numbers (cg1 , cg2 , . . .) indexed by the elements g1 , g2 , . . . of G. On V we define a complex inner product
by
∑
(⃗x, ⃗y ) =
xg y g
g∈G
for any ⃗x = (xg1 , xg2 , . . .), ⃗y = (yg1 , yg2 , . . .) in V . The bar denotes complex
conjugation. We let any h ∈ G act on V by
h⃗x = (xhg1 , xhg2 , . . .).
So we see that any h simply permutes the coordinates of ⃗x in a way prescribed
by the groupstructure. In particular, the elements of G act as unitary (=length
preserving) linear maps on V . Let χ be a character on G. We define the corresponding charactervector ⃗vχ by
⃗vχ = (χ(g1 ), χ(g2 ), . . .)
in other words, ⃗vχ is just the sequence of values of χ. Notice that for any h ∈ G,
h⃗vχ = (χ(hg1 ), χ(hg2 ), . . .) = χ(h)(χ(g1 ), χ(g2 ), . . .) = χ(h)⃗vχ .
Hence ⃗vχ is a common eigenvector for all h ∈ G with eigenvalues χ(h). Conversely, any common eigenvector ⃗v of all h ∈ G defines a character on G by
taking the eigenvalue of each h as charactervalue. Let h⃗v = λ(h)⃗v and suppose we had taken g1 = e. Then h(ve , . . .) = (vh , . . .) and, on the other hand,
h(ve , . . .) = λ(h)(ve , . . .) = (λ(h)ve , . . .). Hence vh = λ(h)ve for all h ∈ G. So
⃗v = ve (λ(g1 ), λ(g2 ), . . .). In particular, ⃗v is uniquely defined up to a scalar factor.
Finally, let χ1 , χ2 be two distinct characters. In particular, χ1 χ2 = χ1 χ−1
2 is not
the trivial character. Then, using Lemma 6.1.3,
∑
∑
(⃗vχ1 , ⃗vχ2 ) =
χ1 (g)χ2 (g) =
(χ1 χ2 )(g) = 0,
g∈G
g∈G
hence ⃗vχ1 and ⃗vχ2 are orthogonal.
Lemma 6.1.4 |Ĝ| = |G|.
F.Beukers, Elementary Number Theory
64
CHAPTER 6. DIRICHLET CHARACTERS AND GAUSS SUMS
Proof. Since, by the above remarks, character vectors belonging to distinct characters are orthogonal, there cannot be more than |G| characters. Suppose that we
have k distinct characters and that k < |G|. Then the orthogonal complement U
in V of the character vectors has dimension |G| − k > 0. Moreover, every element
h ∈ G is unitary and hence maps U into itself. From linear algebra we know
that any set of commuting unitary operators has a common eigenvector, which
in its turn defines a character of G, distinct from the ones we already had. So
we now have k + 1 distinct characters. We continu this procedure untill we have
|G| distinct characters.
2
Lemma 6.1.5 Let G be a finite abelian group and g ∈ G not the identity element.
Then,
∑
χ(g) = 0.
χ∈Ĝ
Proof. Let G = {g1 , g2 , . . .} and Ĝ = {χ1 , χ2 , . . .}. Consider the |G| by |G|
matrix M = (χi (gj ))i,j=1,...,|G| . We have seen in the remarks above that the rows
(letting
j run) of this
√ matrix are orthogonal. Moreover, all rows have lengths
√
|G|. Hence M is |G| times a unitary matrix. In such a matrix the columns
are orthogonal as well, in other words for any distinct g, h ∈ G we have
∑
χ(g)χ(h) = 0.
χ
Our Lemma follows by taking h = e.
2
Definition 6.1.6 A Dirichlet character is a character on (Z/mZ)∗ .
Remark 6.1.7 Let χ be a Dirichlet character on (Z/mZ)∗ . Often one extends
χ to a function on Z by taking the value χ(n) for all n with (n, m) = 1 and
χ(n) = 0 whenever (n, m) > 1. By abuse of language we then still speak of a
Dirichlet character.
Examples
( )
i. The Legendre symbol p. on (Z/pZ)∗ for odd primes p and the Jacobi
symbol on (Z/mZ)∗ for odd m. Both examples are real characters of order
two.
ii. The character of order four on (Z/13Z)∗ given by
a: 1 2 3
4 5 6 7 8 9 10 11 12 (mod 13)
χ(a) : 1 i 1 −1 i i −i −i 1 −1 −i −1
One easily verifies that χ(ab) = χ(a)χ(b) for all a, b ∈ (Z/13Z)∗ . We have
constructed this character by assigning χ(2) = i. Since 2 is a primitive root
modulo 13, the powers of 2 run through (Z/13Z)∗ and we can determine
the character values correspondingly.
F.Beukers, Elementary Number Theory
6.2. GAUSS SUMS, JACOBI SUMS
65
iii. The non-trivial characters χ1 , χ2 , χ3 on (Z/8Z)∗ given by
a:
χ1 (a) :
χ2 (a) :
χ3 (a) :
1
3
5
7
1 −1
1 −1
1 −1 −1
1
1
1 −1 −1
Definition 6.1.8 A Dirichlet character χ is called even if χ(−1) = 1 and odd if
χ(−1) = −1.
6.2
Gauss sums, Jacobi sums
Although it is possible to consider more general Gauss sums, we shall restrict
ourselves here to (Z/pZ)∗ where p is an odd prime. For the p-th roots of unity
we mention the following Lemma. In the sequel we let ζp denote e2πi/p .
Lemma 6.2.1 Let n ∈ Z and p an odd prime. Then,
p−1
∑
{
ζpnx
=
x=1
−1
if p/
|n
p − 1 if p|n
Proof. Clearly, when p|n all terms in the summation are 1, so the second case
follows immediately. In the first case we use the summation formula for geometric
sequences to obtain
(ζppn − ζpn )/(ζpn − 1) = −1.
Definition 6.2.2 Let p be an odd prime and χ a Dirichlet character on (Z/pZ)∗ .
The Gauss sum Sχ is defined by
Sχ =
p−1
∑
χ(x)ζpx .
x=1
Theorem 6.2.3 Let χ be a non-trivial Dirichlet character on (Z/pZ)∗ . Then,
a) Sχ Sχ−1 = χ(−1)p,
b) |Sχ |2 = p,
c) if χ is the Legendre symbol then Sχ2 = (−1)
p−1
2
p.
F.Beukers, Elementary Number Theory
66
CHAPTER 6. DIRICHLET CHARACTERS AND GAUSS SUMS
Proof. Part a) We have
Sχ Sχ−1 =
p−1
∑
χ(x)χ(y)−1 ζpx+y .
x,y=1
Replace x by zy and notice that if z runs through (Z/pZ)∗ then so does x = zy
for any fixed y ∈ (Z/pZ)∗ . Hence
Sχ Sχ−1 =
p−1
∑
χ(z)χ(y)χ(y)−1 ζpy(z+1) .
z,y=1
The factors χ(y) cancel. According to Lemma 6.2.1 summation over y now yields
−1 if z ̸≡ −1(mod p) and p − 1 if z ≡ −1(mod p). Hence,
Sχ Sχ−1 = χ(−1)p −
p−1
∑
χ(x).
x=1
The summation on the right vanishes according to Lemma 6.1.3 and we have
proved assertion a).
Part b) First of all, notice that
Sχ−1 =
p−1
∑
χ(x)−1 ζpx
x=1
p−1
=
∑
(−x)
χ(x)ζp
x=1
= χ(−1)
p−1
∑
(−x)
χ(−x)ζp
x=1
= χ(−1)Sχ .
Hence |Sχ |2 = Sχ Sχ = χ(−1)Sχ Sχ−1 and assertion b) follows by using assertion
a).
c) If χ is the Legendre symbol we have χ = χ−1 and hence, via assertion a),
2
Sχ2 = χ(−1)p = (−1)(p−1)/2 p.
Definition 6.2.4 Let χ1 , χ2 be Dirichlet characters modulo p, where p is an odd
prime. The Jacobi sum J(χ1 , χ2 ) is defined by
J(χ1 , χ2 ) =
p−1
∑
x=2
F.Beukers, Elementary Number Theory
χ1 (x)χ2 (1 − x).
6.3. APPLICATIONS
67
Theorem 6.2.5 Let notations be as in Definition 6.2.4. Suppose that χ1 and χ2
are not each other’s inverse. Then,
Sχ1 Sχ2 = J(χ1 , χ2 )Sχ1 χ2 .
Proof. Write
Sχ1 Sχ2 =
p−1
∑
χ1 (x)χ2 (y)ζpx+y .
x,y=1
We introduce the new summation variables u, v via x = uv, y = u(1 − v).
Conversely, u = x + y, v = x/(x + y). One easily verifies that this gives us a
bijection between the sets {(x, y) ∈ (Z/pZ)∗ 2 | x + y ̸≡ 0(mod p)} and {(u, v) ∈
(Z/pZ)∗ 2 | v ̸≡ 1(mod p)}. Hence,
Sχ1 Sχ2 =
∑
χ1 (x)χ2 (y)ζp(x+y) +
p−1 p−1
∑
∑
x≡−y
u=1 v=2
∑
p−1 p−1
∑
∑
p−1
=
χ1 (uv)χ2 (u(1 − v))ζpu
χ1 (−1)χ1 (x)χ2 (x) +
x=1
χ1 (v)χ2 (1 − v)χ1 (u)χ2 (u)ζpu .
u=1 v=2
The first summation vanishes because of Lemma 6.1.3 and the fact that χ1 χ2 ̸=
χ0 . In the second summation the sum over u yields Sχ1 χ2 and summing over v
yields J(χ1 , χ2 ). Hence Sχ1 Sχ2 = J(χ1 , χ2 )Sχ1 χ2 .
2
Corollary 6.2.6 Let notations be as in Definition 6.2.4. Suppose that the char√
acters are not each other’s inverse. Then |J(χ1 , χ2 )| = p.
Proof. This follows from Theorem 6.2.5 and Theorem 6.2.3(b).
6.3
2
Applications
A nice application of Gauss sums is a short proof of the quadratic reciprocity law.
Let p, q be two odd primes. Write τp = Sχ where χ is the Legendre symbol on
(Z/pZ)∗ . The following calculations will be performed in Z[ζp ] considered modulo
q. Using Theorem 6.2.3(c) and Euler’s Theorem 5.1.3(b) we find that
( )
p−1
q−1
p−1 q−1
p
q−1
q
τp ≡ τp τp ≡ ((−1) 2 p) 2 τp ≡ (−1) 2 2
τp (mod q).
q
On the other hand,
τpq ≡
( p−1 ( )
∑ x
x=1
p
)q
ζpx
F.Beukers, Elementary Number Theory
68
CHAPTER 6. DIRICHLET CHARACTERS AND GAUSS SUMS
≡
p−1 ( )
∑
x
ζpqx
( )∑
)
p−1 (
q
qx
≡
ζpqx
p x=1 p
p
x=1
(
)
q
≡
τp (mod q).
p
Comparison of the above two equalities yields
( )
( )
p−1 q−1
q
p
τp ≡
τp (mod q).
(−1) 2 2
q
p
Multiply on both sides by τp and cancel the resulting τp2 = (−1)
obtain
( ) ( )
p−1 q−1
q
p
(−1) 2 2
≡
(mod q)
q
p
p−1
2
p. We then
and since q ≥ 3, strict equality follows.
( )
Using the above idea we can also compute p2 . The following calculation will
be performed in Z[i] modulo p. We consider (1 + i)p (mod p) and compute it in
two ways. First of all
{
p−1
1 + i(mod p)
if p ≡ 1(mod 4)
p
p
(1 + i) ≡ 1 + i ≡ 1 + (−1) 2 i ≡
−i(1 + i)(mod p) if p ≡ 3(mod 4)
On the other hand, since (1 + i)2 = 2i,
(1 + i) ≡ (2i)
p
p−1
2
( )
2 p−1
(i + 1) ≡
i 2 (1 + i)(mod p).
p
Comparison of these two results shows that
( ) { − p−1
p−1
2
i 2 ≡ (−1) 4
if p ≡ 1(mod 4)
≡
p−1
p+1
−
p
−i · i 2 ≡ (−1) 4 if p ≡ 3(mod 4)
( )
One now easily checks that
±3(mod 8).
2
p
= 1 if p ≡ ±1(mod 8) and
( )
2
p
= −1 if p ≡
Theorem 6.3.1 Let p be a prime with p ≡ 1(mod 4). Then p can be written as
the sum of two squares. More precisely, p = a2 + b2 with
)
2 (
∑
x(x2 − 1)
p−1
a=
x=1
p
)
2 (
∑
x(x2 − ν)
p−1
,
where ν is any quadratic nonresidue .
F.Beukers, Elementary Number Theory
b=
x=1
p
6.3. APPLICATIONS
69
Remark 6.3.2 The sum
cobsthal sum.
∑p−1 ( x(x2 −1) )
x=0
p
, which equals 2a, is known as the Ja-
Proof. Since p ≡ 1(mod 4) we have a character of order 4 on (Z/pZ)∗ which we
call χ4 . This can be constructed by taking χ(g) = i for a primitive root g. Any
other element of a ∈ (Z/pZ)∗ has the form a ≡ g k (mod p) and so( χ(a)
) is fixed( by
)
.
k
k
k
χ(a) = χ(g ) = χ(g) = i . Consider the Jacobi sum J = J(χ4 , p ), where p.
is the Legendre symbol. Notice that J ∈ Z[i], i.e. J = a + bi, a, b ∈ Z. Moreover,
by Corollary 6.2.6, p = |J|2 = a2 + b2 . Hence our first assertion is proved.
Notice that in the summation
J=
p−1
∑
(
χ4 (x)
x=1
1−x
p
)
a term is real if x is a quadratic residue and purely imaginary if x is a quadratic
nonresidue . Hence
(
p−1
a = ℜJ =
2
∑
2
χ4 (y )
y=1
1 − y2
p
)
)
2 ( )(
∑
y
1 − y2
p−1
=
y=1
p
p
)
2 (
∑
y(y 2 − 1)
p−1
=
y=1
p
.
Taking ν to be any quadratic nonresidue we get
(
p−1
ib = iℑJ =
2
∑
2
χ4 (νy )
y=1
1 − νy 2
p
)
)
2 ( )(
∑
y
1 − νy 2
p−1
= χ4 (ν)
y=1
p
p
)
2 (
∑
y(y 2 − ν)
p−1
= −χ4 (ν)
y=1
p
.
Since χ4 (ν) = ±i our asserted value of follows up to ± sign, which is sufficient. 2
Another amusing application is the following theorem discovered by Gauss,
Theorem 6.3.3 Let p be an odd prime for which p ≡ 1(mod 3). Then,
∃x ∈ Z : x3 ≡ 2(mod p) ⇐⇒ ∃A, B ∈ Z : p = A2 + 27B 2 .
When p ≡ −1(mod 3), the congruence x3 ≡ 2(mod p) always has a solution.
F.Beukers, Elementary Number Theory
70
CHAPTER 6. DIRICHLET CHARACTERS AND GAUSS SUMS
Proof. When p ≡ 2(mod 3) we simply take x = 2(2−p)/3 as solution (verify!). So
we suppose that p ≡ 1(mod 3). Then there exists a character χ mod p of order
3. First of all we notice that
∃x ∈ Z : x3 ≡ 2(mod p) ⇐⇒ χ(2) = 1.
Notice that the third roots of unity are all distinct modulo 2. Hence it follows
from Lemma 6.3.4 that
∃x ∈ Z : x3 ≡ 2(mod p) ⇐⇒ J(χ, χ) ≡ 1(mod 2).
We know that J(χ, χ) is an element of Z[ω], the ring of Eisenstein integers, where
ω is a cube root of unity. Thus we have J(χ, χ) = a + bω for some a, b ∈ Z. From
Lemma 6.3.4 it follows that a ≡ −1(mod 3) and b ≡ 0(mod 3). Similarly, if
χ(2) = 1 we find
√ that b ≡ 0(mod 2). When b is√even we can rewrite a + bω as
a − b/2 + (b/2) −3 hence it is an element of Z[ −3]. So we now find
√
∃x ∈ Z : x3 ≡ 2(mod p) ⇐⇒ ∃A, B ∈ Z : J(χ, χ) = A + 3B −3
where we have put A = a − b/2, B = b/6.
√
Suppose that x3 ≡ 2(mod p) has a solution. ThenJ(χ, χ) = A + 3B −3 and
since |J(χ, χ)|2 = p, we find that p = A2 + 27B 2 .
2
Suppose conversely that p can be written as p = A√
+ 27B 2 . Then we have in
Z[ω] the factorisation p = ππ where π = A ± 3B −3. Choosing the proper
sign, we know, by unique factorisation, that π equals J(χ, χ) up to a unit. Since
π ≡ ±1(mod 3) and J(χ,
√ χ) ≡ −1(mod 3) this unit must be ±1 and hence J(χ, χ)
is of the form A′ + 3B ′ −3. This implies that x3 ≡ 2(mod p) is solvable.
Lemma 6.3.4 Let p be an odd prime satisfying p ≡ 1(mod 3). Let χ be a Dirichlet character mod p of order 3. Then,
J(χ, χ) ≡ −1(mod 3)
J(χ, χ) ≡ χ(2)(mod 2).
Proof. According to Theorem 6.2.5 we have
Sχ3 = Sχ Sχ2 J(χ, χ) = pJ(χ, χ).
Notice that modulo 3 we have
Sχ3 ≡
p−1
∑
χ(x)3 ζp3x (mod 3)
x=1
p−1
≡
∑
ζp3x ≡ −1(mod 3)
x=1
where the last equality follows from Lemma 6.2.1.
F.Beukers, Elementary Number Theory
6.4. EXERCISES
71
Notice that modulo 2 we have
Sχ4 ≡
p−1
∑
χ(x)4 ζp4x (mod 2)
x=1
−1
≡ χ(4)
p−1
∑
χ(4x)ζp4x (mod 2)
x=1
≡ χ(2)Sχ (mod 2)
Our congruence mod 2 follows after multiplication by Sχ̄ , Theorem 6.2.3(a) and
p ≡ 1(mod 2).
2
6.4
Exercises
Exercise 6.4.1 Determine all Dirichlet characters modulo 7 and modulo 12.
Exercise 6.4.2 Choose a character χ4 of order 4 and a character χ2 of order 2
on (Z/13Z)∗ . Give a table in which x, χ2 (x), χ4 (x) and χ2 (x)χ4 (1 − x) are listed
for x = 1, . . . , 12. Compute the Jacobi sum J(χ2 , χ4 ). Note that it is a number
in Z[i] and that its norm is 13.
Exercise 6.4.3 Proof that a cyclic group of order N has precisely N characters,
which again form a cyclic group.
Exercise 6.4.4 Let p be an odd prime. Prove that there exists a Dirichlet character of order 4 if and only if p ≡ 1(mod 4).
Exercise 6.4.5 Let p be a prime and a ∈ Z, not divisible by p. Suppose p ≡
2(mod 3). Prove that x3 ≡ a(mod p) has precisely one solution modulo p.
Exercise 6.4.6 Determine all solutions of x3 ≡ 2(mod p) for p = 19, 31. Check
also if p can be written in the form p = a2 + 27b2 .
F.Beukers, Elementary Number Theory
Chapter 7
Sums of squares, Waring’s
problem
7.1
Sums of two squares
In previous chapters we have already seen that if an odd prime divides the sum of
two relatively prime squares, it is 1(mod 4). Moreover, Fermat showed that any
prime ≡ 1(mod 4) can be written as a sum of two squares. Furthermore it was
observed since antiquity that any positive integer can be written as the sum of
four squares. This was proved by Lagrange for the first time in 1770. Problems
such as these form the subject of this section. It should be noted explicitly here
that by a square we mean the square of a number in Z. So, 02 is also considered
to be a square.
The quickest and most elegant way to deal with the above mentioned problems
is to work in the rings of Gaussian integers and quaternionic integers where
we rely heavily on the fact that they are euclidean domains. There exist also
presentations of this subject which work entirely in Z. However, they are in fact
a disguised form of application of the euclidean algorithm.
Theorem 7.1.1 (Fermat) Let p be an odd prime. Then p is a sum of two
squares if and only if p ≡ 1(mod 4).
Proof. Suppose p = x2 + y 2 for some x, y ∈ Z. Since squares are either 0 or 1
mod 4, p, being odd, can only be 1(mod 4).
Now suppose that p ≡ 1(mod 4). Then the congruence equation x2 ≡ −1(mod p)
has a solution, x0 say. Let us now work in Z[i] and use unique factorisation. We
have p|(x20 + 1) hence p|(x0 + i)(x0 − i). If p were prime in Z[i] we would have
p|(x − i) and, via complex conjugation, p|(x + i). Hence p|2i, which is impossible.
Hence p = αβ in Z[i] with N α, N β > 1. Take norms on both sides, p2 = N αN β.
Since N α, N β > 1 this implies p = N α, hence p can be written as a sum of two
squares.
2
72
7.1. SUMS OF TWO SQUARES
73
Theorem 7.1.2 Let r2 (n) be the number of solutions x, y ∈ Z to n = x2 + y 2 .
i. The function r2 (n)/4 is multiplicative.
ii. Let p be a prime. Then

k + 1
r2 (pk )  0
=

4
 1
1
if
if
if
if
p ≡ 1(mod 4)
p ≡ 3(mod 4), k odd
p ≡ 3(mod 4), k even
p=2
Proof. Part i). We note that there is a 1-1-correspondence between solutions of
n = x2 + y 2 and factorisations n = αα in Z[i], simply by taking α = x + yi. Also
note that r2 (n)/4 is precisely the number of factorisations n = αα where we count
αα, iα · −iα, −α · −α, −iα · iα as being the same., i.e. we count factorisations
modulo units.
Let m, n ∈ N and (m, n) = 1. Let mn = λλ, λ ∈ Z[i]. Let µ = gcd(λ, m),
ν = gcd(λ, n). Note that µ, ν are unique up to units. Then λ = µν (up to
units) and mn = µνµν = µµνν. Hence every representation of mn as sum of
two squares corresponds to a pair of representations of m = µµ and n = νν.
Conversely, every such pair gives rise to representation mn = µµνν. Hence
r2 (mn)/4 = (r2 (m)/4)(r2 (n)/4).
Part ii). Let p = 2. Since 2 has only one prime divisor, 1 + i, in Z[i] we have, up
to units, the unique factorisation
2k = (1 + i)k (1 − i)k .
Hence r2 (2k )/4 = 1.
Let p ≡ 3(mod 4). Then p itself is prime in Z[i]. For k odd there is no factorisation of the form pk = λλ, when k is even we have only pk = pk/2 pk/2 .
Let p ≡ 1(mod 4). Then p = ππ, where π, π are distinct primes in Z[i]. For
the factorisation pk = λλ we have, up to units, k + 1 choices, namely λ =
π k , π k−1 π, . . . , π k . Hence r2 (pk )/4 = k + 1, as asserted.
2
In order to solve n = x2 + y 2 for given n there exist several possibilities. When n
is small it is best to solve it by trial and error. When n is larger, there are two
possibilities. For medium sized numbers with several prime factors we suggest the
following method. Let n = pk11 · · · pkr r be the prime factorisation of n. Write each
k
factor pj j as a sum of two squares a2j + b2j . Let αj be one the complex numbers
∏
∏
∏
k
aj ± bj i. Let rj=1 αi = α = a + bi. Notice that n = rj=1 pj j = rj=1 N (αj ) =
N (α) = a2 + b2 . Since we have different choices for each αj we will get different
solutions to n = x2 + y 2 . We have seen that we get essentially all solutions in this
way. As an example we write 65 as sum of two squares. Notice that 65 = 5 × 13
F.Beukers, Elementary Number Theory
74
CHAPTER 7. SUMS OF SQUARES, WARING’S PROBLEM
and 5 = 12 + 22 and 13 = 22 + 32 . The multiplications (1 + 2i)(2 + 3i) = −4 + 7i
and (1 − 2i)(2 + 3i) = 8 − i produce the solutions 65 = 42 + 72 and 65 = 82 + 12 .
According to Theorem 7.1.2 there are 16 solutions and they can be obtained from
the two solutions we found by interchanging x, y and the substitutions x → −x,
y → −y.
When n is a large prime we use a more sophisticated method. Via the techniques
described in the chapter on quadratic reciprocity we first solve l2 ≡ −1(mod n)
and then determine a + bi = gcd(n, l + i) via the euclidean algorithm in Z[i]. We
assert that n = a2 + b2 .
7.2
Sums of more than two squares
Theorem 7.2.1 (Lagrange 1770) Every positive integer can be written as a
sum of four squares.
Proof. First of all notice the equivalenc of the statements
i. ‘n is sum of four squares’
ii. ‘n is the norm of a quaternionic integer’.
Going from i. to ii. is trivial. Suppose conversely that n = N α for some
quaternionic integer α. According to Theorem 13.4.2 there exists a unit ϵ such
that αϵ = p + qi + rj + sk for some p, q, r, s ∈ Z. Hence n = N α = N αϵ =
p2 + q 2 + r2 + s2 .
Secondly, if n, m can be written as a sum of four squares then so can nm. This
follows from N αN β = N αβ. Thus it suffices to prove that any prime p can be
written as a sum of four squares. We assume that p ≥ 5 and proceed as follows.
First we show that there exist x, y ∈ Z such that x2 +y 2 +1 ≡ 0(mod p). Consider
the sets {i2 ∈ Z/pZ| 0 ≤ i ≤ p−1
} and {−1 − i2 ∈ Z/pZ| 0 ≤ i ≤ p−1
}. Each set
2
2
p+1
consists of 2 distinct elements in Z/pZ. Hence these sets overlap, and there
exist x, y such that x2 ≡ −1 − y 2 (mod p), as desired.
By shifting mod p we can see to it that we have x, y ∈ Z such that x2 + y 2 + 1 ≡
0(mod p) and |x|, |y| < p/2. Application of Theorem 13.2.3 to the numbers p and
1 + xi + yj yields a common right-divisor δ ∈ q (the right-‘gcd’) and α, β ∈ q
such that
δ = αp + β(1 + xi + yj).
(7.1)
In the remainder of the proof b|a means: ∃c ∈ q such that a = cb. First of all,
δ|p ⇒ N δ|p2 ⇒ N δ = 1, p or p2 . Secondly, δ|(1 + xi + yj) ⇒ N δ|(1 + x2 + y 2 ) ⇒
N δ < 2 · (p/2)2 + 1 < p2 . Hence N δ = 1 or p. Suppose N δ = 1. Then, according
to Theorem 13.4.4 δ is a unit. Multiply Eq.(7.1) on both sides from the right by
(1 − xi − yj) to obtain δ(1 − xi − yj) = α(1 − xi − yj)p + β(1 + x2 + y 2 ). Since
F.Beukers, Elementary Number Theory
7.2. SUMS OF MORE THAN TWO SQUARES
75
p|(1 + x2 + y 2 ) we infer p|δ(1 − xi − yj) and hence p|(1 − xi − yj), which is clearly
impossible. Thus we conclude that N δ = p and p is a sum of four squares.
2
Not everyone may be charmed by quaternions and for this reason we give Lagrange’s original proof of Theorem 7.2.1 as well.
Proof. Just as in the previous proof we note that it suffices to prove that every
prime p can be written as the sum of four squares. This is a consequence of
Euler’s identity,
(a2 + b2 + c2 + d2 )(a′2 + b′2 + c′2 + d′2 ) =
= (aa′ + bb′ + cc′ + dd′ )2
+(ab′ − ba′ + cd′ − dc′ )2
+(ac′ − bd′ − ca′ + db′ )2
+(ad′ + bc′ − cb′ − da′ )2 .
Just as above, we can find x, y such that x2 +y 2 +1 ≡ 0(mod p) and |x|, |y| ≤ p/2.
We then have x2 + y 2 + 1 = m0 p and m0 ≤ ((p/2)2 + (p/2)2 + 1)/p < p.
Let m be the smallest positive integer such that mp can be written as the sum
of four squares. We will show that m > 1 leads to a contradiction. Suppose
mp = a2 + b2 + c2 + d2
and m > 1.
(7.2)
Choose A, B, C, D in the interval (−m/2, m/2] such that
a ≡ A(mod m), b ≡ B(mod m), c ≡ C(mod m) d ≡ D(mod m).
Then,
A2 + B 2 + C 2 + D2 ≡ a2 + b2 + c2 + d2 ≡ mp ≡ 0(mod m).
Hence,
mr = A2 + B 2 + C 2 + D2 ,
r ≥ 0,
(7.3)
where r = (A2 + B 2 + C 2 + D2 )/m ≤ (4 · (m/2)2 )/m = m. First we show that r ̸=
0, m. If r = 0, then A = B = C = D = 0 and hence m2 |a2 +b2 +c2 +d2 = mp from
which follows m|p, contradicting 1 < m < p. If r = m then A = B = C = D =
m/2 and hence a ≡ b ≡ c ≡ d ≡ m/2(mod m). Notice that if x ≡ m2 (mod m)
then x2 ≡ ( m2 )2 (mod m2 ). And so, mp = a2 +b2 +c2 +d2 ≡ 4(m/2)2 ≡ 0(mod m2 ),
which implies that m2 |mp and thus m|p, again a contradiction. We conclude,
0 < r < m.
(7.4)
Notice that in the proof of (7.4) we have used the assumption m > 1. Now
multiply (7.2) and (7.3) and use Euler’s identity to obtain
mp · mr = (a2 + b2 + c2 + d2 )(A2 + B 2 + C 2 + D2 ) = α2 + β 2 + γ 2 + δ 2
F.Beukers, Elementary Number Theory
76
CHAPTER 7. SUMS OF SQUARES, WARING’S PROBLEM
where
α = aA + bB + cC + dD ≡ a2 + b2 + c2 + d2 ≡ 0(mod m)
β = aB − bA + cD − dC ≡ ab − ab + cd − cd ≡ 0(mod m)
γ = aC − Ac + dB − bD ≡ ac − ac + bd − bd ≡ 0(mod m)
δ = aD − dA + bC − cB ≡ ad − ad + bc − bc ≡ 0(mod m)
So m divides α, β, γ, δ and we find rp = (α/m)2 + (β/m)2 + (γ/m)2 + (δ/m)2 .
We conclude that rp is a sum of four squares and 0 < r < m. This contradicts
the minimality of m. Hence we have m = 1, as desired.
2
For completeness we like to mention the following theorem,
Theorem 7.2.2 (Jacobi) Let n ∈ N. Then the number of solutions to
n = x21 + x22 + x23 + x24 ,
equals
x 1 , x2 , x3 , x4 ∈ Z
∑′
d
8
where the ′ denotes summation over all d with d|n, 4/
|d.
As to the question whether positive integers can be written as a sum of three
squares, we can easily give infinitely many counterexamples.
Theorem 7.2.3 A positive integer of the form 4l (8k + 7) cannot be written as a
sum of three squares.
Proof. We proceed by induction on a. First of all, a number which is 7(mod 8)
cannot be the sum of three squares, simply because the sum of three squares can
never be 7(mod 8). This can easily be verified using the fact that a square can
only be 0, 1 or 4(mod 8).
Suppose l ≥ 0 and suppose we proved our theorem for numbers of the form
4l (8k +7). Let n = 4l+1 (8k + 7) and suppose n = a2 +b2 +c2 . Since n ≡ 0(mod 4)
we have necessarily that a, b, c are all even. Hence n/4 = (a/2)2 + (b/2)2 + (c/2)2 ,
contradicting our induction hypothesis.
2
The converse statement is much harder to prove.
Theorem 7.2.4 (Gauss) Any positive integer not of the form 4l (8k + 7) can be
written as the sum of three squares.
Corollary 7.2.5 Any positive integer can be written as the sum of three triangular numbers.
( )
A triangular number is a number of the form n2 .
F.Beukers, Elementary Number Theory
7.3. THE 15-THEOREM
77
Proof. Let n ∈ N. Write 8n+3 as the sum of three squares, 8n+3 = a2 +b2 +c2 .
Notice that a, b, c all have to be odd. Write a = 2p − 1, b = 2q − 1, c = 2r − 1.
Then 8n + 3 = 4(p2 − p) + 4(q 2 − q) + 4(r2 − r) + 3 and this amounts to
( ) ( ) ( )
p
q
r
n=
+
+
.
2
2
2
(Quoting Gauss: EU REKA : num = ∆ + ∆ + ∆!!)
2
Gauss also proved results on the number of representations of a number as sum
of three squares. As a curiosity we mention
Theorem 7.2.6 Let p be a prime with p ≡ 3(mod 8). Then the number of
solutions to p = x2 + y 2 + z 2 , x, y, z ∈ Z equals
−24 ∑
a
p a=1
p−1
( )
a
.
p
You might recognize the class number h(−p) (see subsection 5.6) in this theorem.
So the number of ways in which a prime p ≡ 3(mod 8) can be written as sum of
three squares equals 24h(−p).
Notice that squares can be considered as quadrangular numbers and Theorem 7.2.1
can be described cryptically as n = 2 + 2 + 2 + 2. Of course we can generalise
this and speak of pentagonal,
( ) hexagonal,... numbers. The general form of the
k-gonal numbers is (k −2) n2 +n. We mention the following theorem, conjectured
by Fermat and proven by Cauchy.
Theorem 7.2.7 (Cauchy) Any positive integer can be written as the sum of k
k-gonal numbers.
7.3
The 15-theorem
Recently Conway and Schneeberger found a remarkable theorem on the representation of integer by quadratic form. An integral quadratic form F (x1 , . . . , xr )
in r variables is a homogeneous polynomial of degree 2 with integer coefficients.
It is called even if the coefficients of the terms xi xj with i ̸= j are all even.
The form F is called positive definite if F (x1 , . . . , xr ) > 0 for all choices of
(x1 , . . . , xr ) ̸= (0, . . . , 0).
We say that a quadratic form F represents an integer if there exist integers
x1 , . . . , xr such that n = F (x1 , . . . , xr ). For example, the form x21 + x22 + x23 + x24
represents all positive integers. In 1993 Conway and Schneeberger announced the
following remarkable theorem.
F.Beukers, Elementary Number Theory
78
CHAPTER 7. SUMS OF SQUARES, WARING’S PROBLEM
Theorem 7.3.1 (Conway, Schneeberger, 1993) An integral, even positive definite quadratic form represents all positive integers if and only if it represents
1, 2, 3, 5, 6, 7, 10, 14, 15.
So, if we want to show that every positive integer can be written in the form
2x2 + 2xy + y 2 + z 2 + u2 with x, y, z, u integers, all we have to do is check whether
each of the numbers 1, 2, 3, 5, 6, 7, 10, 14, 15 can be written in this way. A simple
check shows that this is indeed the case.
The proof of this theorem was quite complicated and never published. However,
in 2000 Manjul Bhargava found a simpler proof and a generalisation to all positive
quadratic forms known as the 290-theorem.
Theorem 7.3.2 (Bhargava, Hanke, 2005) An integral, positive definite quadratic
form represents all positive integers if and only if it represents
1, 2, 3, 5, 6, 7, 10, 13, 14, 15, 17, 19, 21, 22,
23, 26, 29, 30, 31, 34, 35, 37, 42, 58, 93, 110,
145, 203, 290.
7.4
Waring’s problem
Around 1770 Waring put forward the following question . Let k ∈ N and k ≥ 2.
Is there a number g(k) such that any positive integer is the sum of at most g(k)
non-negative k-th powers? About 140 years later, in 1909, Hilbert proved that
the answer is ‘yes’ for any k ≥ 2. His method is quite complicated and we shall
not reproduce it here. Somewhat later Hardy and Littlewood developed a more
conceptual approach to the problem and showed, among other things, that there
exists g(k) ≤ 3 · 2k . From now on we shall denote by g(k) the smallest number
for which Waring’s problem is solvable. The method of Hardy and Littlewood,
and further modifications and improvements of it, are known as circle methods,
which have now become a standard tool in number theory (see R.C.Vaughan,
The circle method). It is now known that g(2) = 4 (Lagrange,1770), g(3) = 9
(Wieferich,Kempner,1912), g(4) = 19 (Balusabramanian, Deshouillers, Dress,
1986), g(5) = 37 (Chen-Jingrun, 1964). For k ≥ 6 we have, if (3/2)k − [(3/2)k ] >
1 − (3/4)k then g(k) = [(3/2)k ] + 2k − 2. It is very likely, but not quite proved
yet, that the inequality always holds.
Although g(3) = 9 it turns out that only 23 and 239 require 9 cubes, all other
numbers can be written as the sum of at most 8 positive cubes. In connection
with this we can define G(k), the smallest integer such that any sufficiently large
integer can be written as the sum of at most G(k) k-th powers. Since infinitely
many numbers are not the sum of three or less squares we have G(2) = 4. It is
also known that 4 ≤ G(3) ≤ 7. It is generally suspected that G(3) = 4 but no
F.Beukers, Elementary Number Theory
7.4. WARING’S PROBLEM
79
one has any idea how to prove this. By a refined version of the circle method
I.M.Vinogradov was able to show that G(k) ≤ 6k(log k +4 +log 216) and Wooley,
in 1992 improved this to G(k) ≤ k log k(1 + o(1)). Special results are G(4) = 16,
G(5) ≤ 17 (conjectured: 6), G(6) ≤ 24 (conjectured: 9), G(7) ≤ 33 (conjectured:
8), G(8) ≤ 42 (conjectured: 32), G(9) ≤ 50 (conjectured: 13), G(10) ≤ 59
(conjectured: 12) (see Vaughan and Wooley, Number Theory for the Milennium
III p301-340 A.K.Peters, 2000).
A variation on Waring’s problem is the following easier problem. Let k ∈ N, k ≥
2. Does there exist a number s such that any n ∈ N can be written in the form
n = ±xk1 ± xk2 ± · · · ± xks ?
If such an s exists we denote its minimal value by v(k). Notice that when k is
odd this question comes down to Waring’s problem for k-th powers of numbers
in Z.
Theorem 7.4.1 We have v(k) ≤ 2k−1 + (k!)/2.
Proof. We introduce the difference operator ∆ which acts on polynomials f (x)
by (∆f )(x) = f (x+1)−f (x). Notice that ∆ decreases the degree of a polynomial
by one. Furthermore,
( ) (
) ( ) (
)
x
x+1
x
x
∆
=
−
=
.
k
k
k
k−1
Notice also that
(
)
( )
x
x
xk
k−1
=
+ r(x)
+
2
k−1
k!
k
where r(x) has degree ≤ k − 2. Application of the operator ∆k−1 yields
( k)
x
k−1
k−1
x+
=∆
.
2
k!
Hence ∆k−1 xk = k!x + k−1
k!. The expression ∆k−1 xk can be considered as the
2
sum of 2k−1 terms of the form ±mk . Now let n ∈ N. Determine x, l ∈ Z such
that
1
k−1
k! + l,
|l| ≤ k!.
n = k!x +
2
2
k−1 k
Hence n = ∆ x + l. Notice that l can be written as the sum of |l| terms 1k or
−1k . Since |l| ≤ (k!)/2 we obtain v(k) ≤ 2k−1 + (k!)/2.
2
The bound of Theorem 7.4.1 is much larger than the actual value of v(k). If we
accept the known upper bounds for G(k) Theorem 7.4.2 yields a much better
result. However, the advantage of Theorem 7.4.1 is that it is simple and selfcontained.
F.Beukers, Elementary Number Theory
80
CHAPTER 7. SUMS OF SQUARES, WARING’S PROBLEM
Theorem 7.4.2 We have v(k) ≤ G(k) + 1.
Proof. Let n ∈ N. Choose y k sufficiently large so that n + y k can be written as
the sum of at most G(k) positive k-th powers, n + y k = xk1 + · · · + xkG(k) . Hence
v(k) ≤ G(k) + 1.
2
Theorem 7.4.3 We have v(2) = 3 and v(3) = 4 or 5.
Proof. Theorem 7.4.1 gives us v(2) ≤ 2 + 1. Notice that also v(2) ≥ 3 because
sums of the form ±a2 ± b2 which are positive can never be 6(mod 8). Hence
v(2) = 3.
Let n ∈ N. Notice that n3 − n is divisible by 6. Write n3 − n = 6x. Notice also
that 6x = (x − 1)3 + (x + 1)3 − 2x3 . Hence n = n3 − (x − 1)3 − (x + 1)3 + 2x3
and we see that v(3) ≤ 5. Moreover, cubes are always 0, ±1(mod 9), so a sum of
three cubes can never be ±4(mod 9). Hence v(3) ≥ 4.
2
It is generally believed, but not proved yet, that v(3) = 4.
Another variation on Waring’s problem is the question, is there a number s such
that any n ∈ N can be written as the sum of s non-negative k-th powers of
rational numbers? I do not know of a simple proof for the existence of such s but
of course the positive solution to Waring’s problem also implies a positive answer
to our question. In fact, we can take s ≤ G(k). For the otherwise elusive case
k = 3 we have a particularly nice solution.
Theorem 7.4.4 (Riley, 1825) Any n ∈ N can be written as the sum of three
positive rational cubes in infinitely many ways.
Proof. Let A = 12t(t + 1) − (t + 1)3 ,
and consider the identity
B = (t + 1)3 − 12t(t − 1),
C = 12t(t − 1)
A3 + B 3 + C 3 = 72t(t + 1)6 .
Let n ∈ N. Choose u ∈ Q such that 1 < nu3 /72 < 2 and take t = nu3 /72. Then
A, B, C, t, t + 1 are all positive and we get
(
A
u(t + 1)2
(
)3
+
B
u(t + 1)2
(
)3
+
C
u(t + 1)2
Moreover, u can be choosen in infinitely many ways.
F.Beukers, Elementary Number Theory
)3
= n.
2
7.5. EXERCISES
7.5
81
Exercises
2
2
Exercise 7.5.1
∑ Let r2 (n) be the number of solutions x, y ∈ Z to n = x + y .
Let R(X) = n≤X r2 (n). Prove that
√
R(X) = πX + O( X),
√
in other words, |R(X) − πX|/ X is bounded (for all
√ X ≥ 1). (Hint: count the
number of lattice points inside the disk with radius X).
Exercise 7.5.2 Find all ways to write 425 as sum of two squares.
Exercise 7.5.3 Which numbers can be written as a difference of two squares?
Exercise 7.5.4 (H.W.Lenstra jr.) Notice, 122 + 332 = 1233, 5882 + 23532 =
5882353. Can you find more of such examples?
Exercise 7.5.5 Let p be a prime number such that p ≡ 1, 3(mod 8).√ Prove
that there exist x, y ∈ Z such that p = x2 + 2y 2 . (Hint: prove that Z[ −2] is
euclidean.)
Exercise 7.5.6 a) Show that 7 cannot be written as a sum of three squares (i.e.
g(2) > 3).
b) Show that 23 and 239 cannot be written as sum of at most 8 cubes (i.e. g(3) >
8).
c) Find N ∈ N such that N cannot be written as a sum of at most 18 fourth
powers (i.e. g(4) > 18).
Exercise 7.5.7 Let g(k) be the function from Waring’s problem. Prove that
g(k) ≥ 2k + [(3/2)k ] − 2. (Hint: consider n = 2k [(3/2)k ] − 1.)
Exercise 7.5.8 We are given the identity
6(a2 + b2 + c2 + d2 )2 = (a + b)4 + (a − b)4 + (c + d)4
+(c − d)4 + (a + c)4 + (a − c)4
+(b + d)4 + (b − d)4 + (a + d)4
+(a − d)4 + (b + c)4 + (b − c)4 .
Prove that g(4) ≤ 53. (Hint: write n = 6N + r and N as sum of four squares.)
Refine the argument to show that g(4) ≤ 50.
Exercise 7.5.9 Show that 5 cannot be written as the sum of 4 fourth powers of
rational numbers. (Hint: look modulo 5).
Can you find other numbers that cannot be written as sum of 4 rational fourth
powers?
F.Beukers, Elementary Number Theory
Chapter 8
Continued fractions
8.1
Introduction
Let α ∈ R. The continued fraction algorithm for α runs as follows.
x0 = α
a0 = [x0 ],
x1 = 1/{x0 }
a1 = [x1 ],
x2 = 1/{x1 }
......
xn+1 = 1/{xn }
an = [xn ],
......
Notice that xi ≥ 1 for all i ≥ 1. The algorithm is said to terminate if {xn } = 0
for some n. Notice that
α = a0 +
1
1
= a0 +
x1
a1 +
1
x2
= a0 +
1
1
a1 + a2 +···
which is denoted as
α = [a0 , x1 ] = [a0 , a1 , x2 ] = [a0 , a1 , a2 , . . . ] .
Theorem 8.1.1 The continued fraction algorithm terminates if and only if α ∈
Q.
Proof. Suppose we have termination, i.e. {xn } = 0 for some n. Then α =
[a0 , a1 , . . . , an−1 ] and we see trivially that α ∈ Q.
If α ∈ Q then the xi are all rational numbers, say xi = pi /qi with pi , qi ∈ N and
pi > qi for all i. Notice that qi+1 = pi − [pi /qi ]qi for all i, hence q1 > q2 > q3 >
· · · > 0. So we see that the algorithm terminates.
2
82
8.1. INTRODUCTION
83
In fact, when α is rational, α = p/q then the continued fraction algorithm is
nothing but the Euclidean algorithm applied to p, q.
Theorem 8.1.2 Let a0 , a1 , . . . , an ∈ R. Suppose
p−2 = 0
q−2 = 1
p−1 = 1 p0 = a0
q−1 = 0 q0 = 1
pn = an pn−1 + pn−2
qn = an qn−1 + qn−2
Then,
[a0 , a1 , . . . , an ] =
(n ≥ 0)
(n ≥ 0)
pn
.
qn
Proof. By induction on n we shall show that [a0 , . . . , an ] = (an pn−1 +pn−2 )/(an qn−1 +
qn−2 ). For n = 0 this is trivial. Now suppose n ≥ 0. Notice that
[a0 , a1 , . . . , an , an+1 ] = [a0 , a1 , . . . , an−1 , an +
1
]
an+1
(an + 1/an+1 )pn−1 + pn−2
=
(an + 1/an+1 )qn−1 + qn−2
an+1 (an pn−1 + pn−2 ) + pn−1
=
an+1 (an qn−1 + qn−2 ) + qn−1
an+1 pn + pn−1
=
an+1 qn + qn−1
2
which completes our induction step.
From now on we shall adhere to the notations α = [a0 , a1 , . . . ] , [a0 , a1 , . . . , an ] =
pn /qn for the continued fraction expansion of α. We call a0 , a1 , a2 , . . . the partial
fractions of the continued fraction and the pn /qn the convergents . Why the pn /qn
are called convergents will become clear from the following theorem.
Theorem 8.1.3 Let notation be as above. Then, for all n ≥ 0,
1.
pn−1 qn − pn qn−1 = (−1)n .
2.
pn
(−1)n
α−
=
.
qn
qn (xn+1 qn + qn−1 )
Proof. Part (1) is proved by induction on n, the case n = 0 being trivial.
pn pn+1 pn an+1 pn + pn−1 =
pn qn+1 − pn+1 qn = qn qn+1 qn an+1 qn + qn−1 pn−1 pn pn pn−1 n
n+1
= −
= qn−1 qn = −(−1) = (−1)
qn qn−1 Part (2) follows from α = [a0 , a1 , . . . , an , xn+1 ] =
ward computation of the difference α − pqnn .
xn+1 pn +pn−1
xn+1 qn +qn−1
and a straightfor2
F.Beukers, Elementary Number Theory
84
CHAPTER 8. CONTINUED FRACTIONS
Corollary 8.1.4 For all convergents pn /qn we have
p
1
n
α − < 1 <
qn
qn+1 qn
an+1 qn2
We see from the previous corollary that convergents p/q of the continued fraction
of an irrational number α have the property that
p
− α < 1
(8.1)
q
q2
In particular this means the convergent give very good rational approximations
with respect to their denominator. As an example consider
π = [3, 7, 15, 1, 292, 1, 1, 1, 2, 1, . . . ].
From the theory we expect that |π − p/q| < 1/(292q 2 ) where p/q = [3, 7, 15, 1]
which equals 355/113. In fact,
π−
355
= −0.000000266764
113
and 355/113 approximates π up to 6 decimal places. Note that there does not
seem to be any regularity in the continued fraction of π.
Here are some other examples of continued fraction expansions:
e = [2, 1, 2, 1, 1, 4, 1, 1, 6, 1, 1, 8, 1, 1, 10, . . . ]
e2 = [7, 2, 1, 1, 3, 18, 5, 1, 1, 6, 30, 8, 1, 1, 9, 42, 11, 1, 1, . . . ]
e3 = [20, 11, 1, 2, 4, 3, 1, 5, 1, 2, 16, 1, 1, 16, 2, 13, 14, 4, 6, 2, 1, 1, 2, 2, . . . ]
√
2 = [1, 2, 2, 2, 2, 2, 2, . . . ]
√
97 = [9, 1, 5, 1, 1, 1, 1, 1, 1, 5, 1, 18, 1, 5, 1, 1, 1, 1, 1, 1, 5, 1, 18, 1, . . . ]
√
47 = [6, 1, 5, 1, 12, 1, 5, 1, 12, 1, 5, 1, 12, 1, 5, 1, . . . ]
√
2
It is interesting to note the regularity in the expansions of e, e√
and N , but
not in e3 . We shall reurn to the periodicity of the expansion of N in the next
section.
It also turns out that if a fraction p/q satisfies (8.1) then it is almost a convergent
of the continued fraction of α.
Theorem 8.1.5 (Legendre) Suppose α ∈ R and p, q ∈ Z, q > 0 such that
p
α − < 1 .
q 2q 2
Then p/q is a convergent of the continued fraction of α
F.Beukers, Elementary Number Theory
8.2. CONTINUED FRACTIONS FOR QUADRATIC IRRATIONALS
85
Proof. Let p0 /q0 , p1 /q1 , . . . be the convergents of the continued fraction of α.
Choose n such that qn ≤ q < qn+1 . Let us assume p/q ̸= pn /qn , otherwise we are
done. Then we have the following inequalities,
p pn 1
1
pn p 1
≤ − ≤ α − + α − <
+ 2
qqn
q qn
qn
q
qn qn+1 2q
q
Multiply these inequalities with qqn to obtain 1 < q2qn + qn+1
. When qn+1 ≥ 2q we
find, using qn ≤ q that 1 < 1/2 + 1/2 which is a contradiction.
So let us assume qn+1 < 2q. We now repeat our estimates with a little more
care. Suppose first that α − p/q and α − pn /qn have the same sign. Then the
absolute value of their difference, which equals |p/q − pn /qn |, is bounded above
by max(1/2q 2 , 1/qn qn+1 ). It is bounded below by 1/qqn . Multiplication by qqn
yields 1 < max(qn /2q, q/qn+1 ) < max(1/2, 1) = 1, again a contradiction.
Now suppose that α−p/q and α−pn /qn have opposite sign. Then α−p/q and α−
pn+1 /qn+1 have the same sign. Just as above we derive 1 < max(qn+1 /2q, q/qn+2 ).
Using qn+1 < 2q we find 1 < max(1, 1) = 1, again a contradiction.
2
Here is a very useful lemma in all that follows.
Lemma 8.1.6 Let α = [a0 , a1 , a2 , . . . , am , β]. Then −1/β = [am , . . . , a2 , a1 , a0 , −1/α].
Proof. This goes by induction on m. For m = 0 the lemma is clear,
α = [a0 , β] ⇒ α = a0 +
1
1
1
1
1
⇒ − = a0 +
⇒ − = [a0 , − ].
β
β
−1/α
β
α
Suppose m > 0. Then, α = [a0 , . . . , am−1 , am +1/β]. By the induction hypothesis
we obtain
1
1
−
= [am−1 , . . . , a1 , a0 , − ]
am + 1/β
α
Invert on both sides and add am to obtain
1
1
− = [am , . . . , a2 , a1 , a0 , − ]
β
α
2
8.2
Continued fractions for quadratic irrationals
A real, non-rational number α which satisfies a polynomial equation of degree
2 over Q is called a quadratic irrational . Given a quadratic irrational there
exist, up to common sign change, a unique triple of integers A, B, C such that
Aα2 +Bα+C = 0 and gcd(A, B, C) = 1, AC ̸= 0. The polynomial AX 2 +BX +C
2
is called the minimal polynomial
√ is called the
√ of α. The number D = B − 4AC
discrminant of α. If α = a + b D for some a, b ∈ Q we call a − b D its conjugate
and denote it by α. A quadratic irrational α is called reduced if α > 1 and
−1 < α < 0.
F.Beukers, Elementary Number Theory
86
CHAPTER 8. CONTINUED FRACTIONS
Theorem 8.2.1 Let α be a quadratic irrational of discriminant D. Put x0 = α
and define recursively xn+1 = 1/(xn − [xn ]). Then each xn has discriminant D
and there exists n0 such that xn is reduced for all n > n0 .
Proof. That the discrimant does not change is clear if we realise that α and
α − m have the samen discriminant for any m ∈ Z and that α and 1/α have the
same discrminant.
Denote by xn the conjugate of xn . Notice that xn > 1 for all n ≥ 1. Verify also
that if xn is reduced the same holds for xn+1 , xn+2 , . . . .
Let m be the smallest index such that [xm ] ̸= [xm ]. If such an m would not exist,
both α and α have the same continued fraction expansion.
Suppose that [xm ] < [xm ]. Then notice that xm+1 = 1/(xm − [xm ]) < 0. From
xm+2 = 1/(xm+1 − [xm+1 ]) and xm+1 < 0 we then conclude that −1 < xm+2 < 0,
hence xm+2 is reduced.
Now suppose that [xm ] > [xm ]. Then xm+1 = 1/(xm − [xm ]) < 1. Hence [xm+1 ] =
0 < [xm+1 ] and we continue as in the preceding case.
2
Theorem 8.2.2 There exist finitely many reduced quadratic irrationals of given
discriminant D.
√
with P, Q ∈ Z,
Proof. Let α be such a quadratic irrational and write α = D+P
Q
√
Q ̸= 0 (we take D > 0.
From α > 0√> α it follows that Q > 0. From α > 1 > −α it follows
√
√ that
D + P >√ D − P , hence P > 0. From α <
0
it
follows
that
P
−
√
√ D < 0,
hence P < D. From α > 1 we conclude
P + D > Q, hence
Q < 2 D.
√
√
Concluding, we find that 0 < P < D and 0 < Q < 2 D, hence we have at
most finitely many possibilities.
2
Let [a0 , a1 , a2 , . . .] be the continued fraction expansion of a real number. We say
that the expansion is periodic if there exist n0 ∈ Z, N ∈ N such that an+N = an
for all n ≥ n0 . We call the expansion purely periodic if n0 = 0.
Theorem 8.2.3 Let α ∈ R. Then the continued fraction expansion of α is
periodic if and only if α is a quadratic irrational. It is purely periodic if and only
if α is reduced.
Proof. We first prove our theorem for purely periodic expansions. Suppose α
has a purely periodic continued fraction. Then there exists an r such that α =
[a0 , a1 , . . . , ar , α]. Hence
αpr + pr−1
α=
αqr + qr−1
This implies qr α2 +(qr−1 −pr )α−pr−1 = 0. First of all we see that α is a quadratic
irrational. Since its continued fraction is purely periodic we must have a0 ≥ 1,
F.Beukers, Elementary Number Theory
8.2. CONTINUED FRACTIONS FOR QUADRATIC IRRATIONALS
87
hence α > 1. From the quadratic equation we see that αα = −pr−1 /qr . Using
either pr−1 /qr = (pr−1 /qr−1 )(qr−1 /qr ) or pr−1 /qr = (pr−1 /pr )(pr /qr ) we conclude
that pr−1 /qr < α, hence −1 < α < 0. So α is reduced.
Suppose conversely that α is a reduced quadratic irrational. Let x0 = α and
recursively xn+1 = 1/(xn −[xn ]). Since x0 is reduced, all xi are reduced. Moreover
their discriminants are all the same, hence there exist only finitely many distinct
xi . So there exist r < s such that xr = xs . Notice that the value of an follows
uniquely from xn+1 by the condition that xn is reduced, namely an = [−1/xn+1 ].
Hence xn = [−1/xn+1 ] + 1/xn+1 . In particular it follows from xr = xs that
xr−1 = xs−1 , etcetera, hence x0 = xs−r . So the continued fraction of x0 = α is
purely periodic.
Now suppose that α has a periodic continued fraction. Then there exists β with
a purely periodic expansion such that α = [a0 , a1 , . . . , an0 , β]. We know that β
is a quadratic irrational from the above, hence the same holds for α. Suppose
conversely that α is a quadratic irrational. Then we know that there is a reduced
β such that α = [a0 , . . . , an0 , β]. Since β has periodic continued fraction, the
same holds for α.
2
√
For quadratic irrational numbers of the form N , N ∈ N not a square, we obtain
the following theorem.
√
Theorem 8.2.4 Let N ∈ N √
and suppose N is not a square. Then N =
[a0 , a1 , . . . , ar , 2a0 ] where a0 = [ N ]. Moreover, (a1 , a2 , . . . , ar ) = (ar , . . . , a2 , a1 ).
√
Proof. First observe that a0 = [ N
√] is the result of the first step in the continued
fraction algorithm. Now note
√
√ that N + a0 is reduced quadratic irrational, since
N + a0 > 1 and −1 < − N + a0 < 0. Hence it has a purely periodic continued
fraction of the form
√
N + a0 = [2a0 , a1 , a2 , . . . , ar ] = [2a0 , a1 , a2 , . . . , ar , 2a0 ].
√
After subtraction of a0 on √
both sides we obtain N = [a
√0 , a1 , . . . , ar , 2a0 ], as
asserted. Notice also that√ N + a0 = [2a0 , a1 , a2 , . . .√
, ar , N + a0 ]. Substract
2a0 on both sides to find N − a0 = [0, a1 , a2 , . . . , ar , N + a0 ]. Hence
√
√
1
= [a1 , a2 , . . . , ar , N + a0 ].
N − a0
Application of Lemma 8.1.6 yields
√
1
= [ar , . . . , a2 , a1 , a0 − N ]
N + a0
√
√
This algebraic identity remains true if we replace N by − N ,
−√
√
1
− √
= [ar , . . . , a2 , a1 , a0 + N ].
− N + a0
F.Beukers, Elementary Number Theory
88
CHAPTER 8. CONTINUED FRACTIONS
Invert both sider and add 2a0 to obtain
√
√
N + a0 = [2a0 , ar , . . . , a2 , a − 1, N + a0 ].
√
So we see that the continued fraction of N +a0 is also given by [2a0 , ar , . . . , a2 , a1 ].
Hence (a1 , a2 , . . . , ar ) = (ar , . . . , a2 , a1 ).
2
8.3
Pell’s equation
Suppose N ∈ N is not a square and consider the diophantine equation
x2 − N y 2 = 1
in the unknowns x, y ∈ Z≥0 . Although problems related to his equation have
been around since antiquity, the first general method for solving it was given
by W.Brouncker in 1657. He was able to use his method to obtain the smallest
solution
(x, y) = (32188120829134849, 1819380158564160)
to x2 − 313y 2 = 1! Brouncker’s method was described in Wallis’s book on algebra and number theory. Euler mistakenly assumed from Wallis’s book that the
method was due to John Pell, another English mathematician. Very soon Pell’s
name stuck to this equation. For several values of N we list the solution with
minimal x,
32 − 2 · 22 = 1
6492 − 13 · 1802 = 1
17663190492 − 61 · 2261539802 = 1
Looking at these examples one observes that it is quite a miracle that any nontrivial solution for x2 − 61y 2 = 1 exists. Nevertheless, using continued fractions
it is possible to show that there always exists a non-trivial solution.
Proposition 8.3.1 Let N ∈ N and suppose N is not a square. Then there exist
x, y ∈ N such that x2 − N y 2 = 1.
Proof. For N = 2, 3, 5, 6 our theorem is true since we have 32 − 2 · 22 = 1,
22 − 3 · 12 = 1, 92 − 5 · 42 = 1, 52 − 6 · 22 = 1.√So we can assume that N ≥ 7.
Consider the continued fraction expansion of N given by
√
N = [a0 , a1 , . . . , ar , 2a0 ]
say. Let p/q = [a0 , a1 , . . . , ar ]. Then, from our elementary estimates we find that
p √ − N < 1 .
2a0 q 2
q
F.Beukers, Elementary Number Theory
8.3. PELL’S EQUATION
89
√
√
Multiply
on
both
sides
by
|p/q
+
N
|
and
use
the
fact
that
|p/q
+
N| ≤
√
(2 N + 1). We find,
√
2
p
− N < 2 N + 1.
q2
2a0 q 2
√
√
Multiply on both sides by q 2 to find |p2 − N q 2 | < (2 N + 1)/2[ N ]. When
N ≥ 7 we have
√
√
2 N +1
2 N +1
√
< √
<2
2[ N ]
2( N − 1)
Hence |p2 −N q 2 | < 2. So we have either p2 −N q 2 = −1 or p2 −N q 2 = 1. (why can’t
we have p2 − N q 2 = 0?). In case p2 − N q 2 = 1 we find x = p, y = q as solution.
In case p2 − N q 2 = −1 we notice that (p2 + N q 2 )2 − N (2pq)2 = (p2 − N q 2 )2 = 1.
Hence we have the solution x = p2 + N q 2 , y = 2pq.
2
Now that we established the existence of non-trivial solutions to Pell’s equation
we would like to have the full set. An important remark to this end is the following
trick which
by an example. Notice that 32 − 2 · 22 = 1 is equivalent
√
√ we illustrate
to (3 + 2 2)(3 − 2 2) = 1. Take the square on both sides and use the fact that
√
√
(3 ± 2 2)2 = 17 ± 12 2
√
√
Hence (17 + 12 2)(17
172 − 2 · 122 = 1. We can also take
√ − 12 2) which implies
√
the cube of (3 + 2 2) to obtain 99 + 70 2. We then find 992 − 2 · 702 = 1. So,
given one solution of Pell’s equation we can construct infinitely many! If we start
with the smallest positive solution we get all solutions in this way, as shown in
the following theorem.
√
Theorem 8.3.2 Choose the solution of Pell’s equation with x + y N > 1 and
minimal. Call it (p, q). Then,
x, y ∈ N of Pell’s equation there
√ to any solution
√
exists n ∈ N such that x + y N = (p + q N )n .
√
2
2
Proof.
√ Notice that if u, v ∈ Z satisfy
√ −1u − N v = 1 and u + v N ≥ 1, then
u − v N , being √
equal to (u + v N ) √
lies between 0 and 1. Addition of the
inequalities u + v N ≥ 1 and 1 ≤ u − v N √
> 0 implies u ≥ 0. Substraction of
these inequalities yields v > 0. We call u + v N the size of the solution
√ u, v.
Now
√ let x, y ∈ N be any solution√of Pell’s equation. Notice that (x + y N )(p −
q N ) = (px − qyN ) + (py
√ − qx) N . Let
√ u = px −√qyN, v = py − qx and we have
2
2
u − N v = 1 and u + v N = (x + y N )/(p + q N ). Observe that
√
√
x+y N
√ < (x + y N )/2,
1≤
p+q N
√
√
hence 1 <= u + v N < (x + y N )/2. So we have found a√new solution with
positive coordinates and size bounded by half the size of x + y N . By repeatedly
F.Beukers, Elementary Number Theory
90
CHAPTER 8. CONTINUED FRACTIONS
performing
this operation we obtain a solution whose size is less than the size of
√
p + q N . By the minimality of p, q this implies that this last solution √
should
be 1, 0.
the number of steps is n we thus find that x + y N =
√ Supposing
n
2
(p + q N ) .
In the existence
√ proof for solutions to Pell’s equation we have used the continued
fraction of N . It turns out that we can use this algorithm to find the smallest
solution and also all solutions of other equations of the form x2 − N y 2 = k for
small k.
√
Theorem 8.3.3 Suppose we have x, y ∈ N such√that |x2 − N y 2 | ≤ N . Then
x/y is a convergent to the continued fraction of N .
√
√
Proof. Let M = [ N ]. Since x2 −N y 2 is integral the inequality |x2 −N y 2 | < N
implies |x2 − N y 2 | ≤ M . We first show that x ≥ M y. If x < M y we would have
the following sequence of inequalities,
x2 − N y 2 < x2 − M 2 y 2 = (x − yM )(x + yM ) < −M
contradicting |x2 − N y 2 | ≤ M . So we have x ≥ M y.
Notice that |x2 − N y 2 | ≤ M implies
√
|x − y N | ≤
M
M
M
1
√ <
≤
=
x + yM
2yM
2y
x+y N
Divide by y on both sides and use Theorem 8.1.5 to conclude that x/y is a
convergent.
2
8.4
Archimedes’s Cattle Problem
The following story has been taken from Albert H.Beiler, Recreations in the
Theory of Numbers, Dover. It deals with a problem which is attributed to
Archimedes. In the form of an epigram one is asked for the number of oxen
of the sun. The oxen go in four colors, white, black, spotted and yellow. Let
W, X, Y, Z be the number of bulls of color white, balck, spotted, yellow respectively and x, y, z, w the number of cows. It is asked that
9
X = 20
Y +Z
W = 65 X + Z
7
Y = 13
W
+
Z
w
=
(X
+ x)
42
12
11
9
x = 20 (Y + y) y = 30 (Z + z)
13
z = 42
(W + w)
and in addition W + X should be a square and Y + Z a triangular number (i.e.
of the form n(n + 1)/2. The first seven conditions are easy and one verifies that
F.Beukers, Elementary Number Theory
8.5. CORNACCHIA’S ALGORITHM
91
there exists k ∈ N such that
W = 10366482k
X = 7460514k
Y = 7358060k
Z = 4149387k
w = 7206360k
x = 4893246k
y = 3515820k
z = 5439213k
The condition W + X a square implies that 17826996k must be a square. Hence
k = 4456749t2 for some integer t. Finally the condition Y + Z = n(n + 1)/2
implies
51285802909803t2 = n(n + 1)/2
Multiply by 8 on both sides and add 1 to obtain
4n2 + 4n + 1 = (2n + 1)2 = 410286423278424t2 + 1
= 4729494(9314t)2 + 1
In other words we are looking for the solutions of the pellian equation
u2 − 4729494v 2 = 1
where v must be divisible by 9314. Finding the smallest solution has only been
possible by the use of a computer and the smallest solution value of u turns out
to have 206554 decimal digits.
8.5
Cornacchia’s algorithm
Closely related to the previous section is the problem to write a prime number in
the form x2 + dy 2 , where d ∈ N is given. It is known for example, that p is a sum
of two squares if and only if p ≡ 1(mod 4). The question is how to find the two
squares if p ≡ 1(mod 4). One approach would be as follows. Determine z ∈ N
such that z 2 ≡ −d(mod p) by the method of the previous section. Then consider
the lattice L generated by the vectors (z, 1) and (p, 0). Moreover, for every (a, b) ∈
L we have a2 + db2 ≡ 0(mod p). Conversely, suppose that a2 + db2 = p is solvable
in integers a, b. Then it follows from a2 ≡ −db2 (mod p) that a ≡ ±zb(mod p).
Hence either (a, b) or (−a, b) is in the lattice L. Suppose (a, b) ∈ L. If we take the
norm x2 + dy 2 on L then (a, b) is a shortest vector in L and we can use reduction
in dimension 2 to find this shortest vector.
However, the reduction algorithm with this particular application can be reformulated in an even simpler way.
Theorem 8.5.1 (G.Cornacchia,1908) Let d ∈ N and p a given odd prime.
Suppose a2 + db2 = p is solvable in the positive integers a, b. If d = 1 we assume
that a > b. Let x0 ∈ N be such that x20 ≡ −d(mod p) and x0 < p/2. Apply the
following algorithm recursively, x1 = p−[p/x0 ]x0 , x2 = x0 −[x0 /x1 ]x1 , . . . , xi+1 =
√
xi−1 − [xi−1 /xi ]xi , . . . . Choose i such that xi < p < xi−1 . Then xi = a.
F.Beukers, Elementary Number Theory
92
CHAPTER 8. CONTINUED FRACTIONS
Proof. Suppose a0 , b0 ∈ N such that a20 + db20 = p and suppose a0 > b0 if d = 1.
Note that (a0 , b0 ), (db0 , −a0 ) span a lattice L of determinant p. Note also that
if (x, y) = λ(a0 , b0 ) + µ(db0 , −a0 ), then x2 + dy 2 = (λ2 + dµ2 )p. In particular,
(x, y) ∈ L ⇒ x2 + dy 2 ≡ 0(mod p).
Let (a1 , b1 ) = (db0 , −a0 ) + [a0 /b0 ](a0 , b0 ) and define recursively
(ai+1 , bi+1 ) = (ai−1 , bi−1 ) + [|bi−1 /bi |](ai , bi ),
i ≥ 1.
Notice that a0 < a1 < a2 < · · · and that |b0 | > |b1 | > |b2 | > · · · and that
(−1)i bi > 0 as long as bi ̸= 0. Notice also that, given ai+1 and ai , one can recover
ai−1 by the conditions ai−1 ≡ ai+1 (mod ai ) and 0 < ai−1 < ai .
Suppose that bk = 0, which will indeed happen for some k. The algorithm
then
terminates.
Since each pair (ai , bi ), (ai−1 , bi−1 ) forms a basis of L, we have
ai−1 bi−1 = ±p. In particular when i = k we obtain ak bk−1 = ±p. Hence
ai
bi ak divides p. On the other hand, 0 ≡ a2k + db2k ≡ a2k (mod p), hence p divides
ak . So we get ak = p and bk−1 = ±1. Hence, from ak−1 + db2k−1 ≡ 0(mod p)
we get a2k−1 ≡ −d(mod p). Starting with the values ak , ak−1 we can compute
ak−2 , ak−3 , . . . recursively. If ak−1 = x0 this will be exactly the recursion procedure
of our theorem. If ak−1 = p−x0 , we note that ak−2 = x0 , ak−3 = p−x0 (mod x0 ) =
p(mod x0 ) and the recursion again coincides with the recursion of our theorem.
√
√
We have trivially a0 < p. If we can show that a1 > p we know exactly at
which point in the euclidean algorithm for p, x0 we have
√ hit upon the desired
√
a0 . To show that a1 > p notice that |b1 | < b0 | < p/d. Moreover, a short
computation shows that a1 + db21 = ([a0 /b0 ]2 + d)p. Using b21 < p/d this implies
a21 > ([a0 /b0 ]2 + d − 1)p. When d > 1 we are done. If d = 1 we know that a0 > b0 ,
hence [a0 /b0 ] ≥ 1 and we are also done.
2
In the case d = 1 the algorithm has some nice side properties. Suppose that
x0 ≡ −1(mod p) and 0 < x0 < p/2. Then the continued fraction algorithm
of p/x0 is symmetric of even length. Moreover, let a and b be the first two
√
remainders less than p in the euclidean algorithm for p/x0 . Then a2 + b2 = p.
All this, and more, is shown in the exercises below.
First we treat the example of writing the prime p = 1020 + 129 as sum of two
squares. Note that m := (p − 1)/27 is odd. We first determine a generator g of
the group of elements modulo p whose order divides 27 . Then, noting that any
solution x of x2 ≡ −1(mod p) has order 4, we simply take x ≡ g 32( as) solution.
( )
The smallest quadratic nonresidue modulo p is 7. We easily check p7 = p7 =
(5)
= −1. Then a generator g can be found by setting g ≡ 7m (mod p), and a
7
solution of x2 ≡ −1(mod p) by x ≡ 732m ≡ 44237909966037281987(mod p). The
continued fraction of p/x equals
[2, 3, 1, 5, 5, 167, 3, 14, 69, 33, 2, 2, 33, 69, 14, 3, 167, 5, 5, 1, 3, 2]
F.Beukers, Elementary Number Theory
8.6. EXERCISES
93
and Cornacchia’s algorithm yields after 11 steps,
p = 89708788732 + 44185215002 .
8.6
Exercises
A small point to be recalled in the following exercises is that the continued fraction
of a rational number is not unique. If we would have p/q = [a0 , . . . , ar , 1] for
example, we can rewrite it as p/q = [a0 , . . . , ar +1]. This would be the result of the
continued fraction algorithm. In the exercises we shall use the word normalised
continued fraction if we want the last partial quotient larger than 1.
Exercise 8.6.1 Let p, q ∈ N be relatively prime and such that q < p. Denote the
continued fraction of p/q by [a0 , . . . , ar ]. Let q ′ be such that qq ′ ≡ (−1)r (mod p).
Using Lemma 8.1.6, show that p/q ′ = [ar , . . . , a0 ].
Exercise 8.6.2 Suppose the rational number p/q has a symmetric continued
fraction, i.e. p/q = [a0 , a1 , . . . , am ] with (a0 , a1 , . . . , am ) = (am , . . . , a1 , a0 ). Using
Lemma 8.1.6, show that q 2 ≡ (−1)m (mod p).
Exercise 8.6.3 Let p, q ∈ N such that gcd(p, q) = 1, q < p/2 and q 2 ≡ ±(mod p).
Show that the normalised continued fraction expansion of p/q is symmetric.
Exercise 8.6.4 Let b/a = [a0 , a1 , . . . , ar ] where ai ∈ N for all i and gcd(a, b) = 1.
Let p/q = [ar , ar−1 , . . . , a0 , a0 , . . . , ar ]. Using Lemma 8.1.6 show that p = a2 + b2 .
F.Beukers, Elementary Number Theory
Chapter 9
Diophantine equations
9.1
General remarks
Let F (x1 , . . . , xr ) ∈ Z[x1 , . . . , xr ]. An equation of the form
F (x1 , . . . , xr ) = 0
(9.1)
in the unknowns x1 , . . . , xr in Z or Q is called a diophantine equation.
Examples are:
x7 + y 7 = z 7 ,
x2 − 67y 2 = 1,
y 2 = x3 − 2.
The behaviour of the solution sets seems to be very erratic and depends very
strongly on small variations of the coefficients. For example x3 +y 3 = z 3 and x3 +
y 3 = 4z 3 are known to have no integral solutions with xyz ̸= 0 whereas x3 + y 3 =
13z 3 and x3 + y 3 = 22z 3 have infinitely many with xyz ̸= 0 and (x, y, z) = 1 (the
‘smallest’ solutions being (x, y, z) = (2, 7, 3), (25469, 17299, 9954) respectively).
A famous problem on diophantine equations was Hilbert’s tenth problem : Is
there a computer program, using unlimited memory, with which one can decide
whether any equation of the form (9.1) has a solution or not. We use the term
’computer program’ here to avoid having to explain the definition of ’algorithm’.
The answer to this question, given in 1970 by Matijasevich is no. The proof is
based on a combination of logic and number theory and unfortunately falls outside
the scope of these notes. The result of Matijasevich suggests that any diophantine
equation has its own peculiarities. On the one hand it makes the solution of
diophantine equations harder, but on the other hand also more interesting because
of the variety of approaches which are now required. In the following sections we
shall discuss a few classes of diophantine equations.
9.2
Pythagorean triplets
One of the ancient diophantine equations is the following.
94
9.2. PYTHAGOREAN TRIPLETS
95
Definition 9.2.1 A triplet a, b, c ∈ N is called Pythagorean if
gcd(a, b, c) = 1
and
a2 + b2 = c2 .
Notice that in a Pythagorean triplet a and b cannot be both odd. For then
we would have a2 + b2 ≡ 1 + 1 ≡ 2(mod 4) but c2 , being a square, cannot be
≡ 2(mod 4).
Theorem 9.2.2 Let a, b, c ∈ N and suppose that b is even. Then a, b, c is a
Pythagorean triplet if and only if ∃r, s ∈ N : r > s, r ̸≡ s(mod 2), (r, s) =
1, a = r2 − s2 , b = 2rs, c = r2 + s2 .
Proof. Suppose we have r, s with the given properties. Clearly a, b, c satisfy
a2 + b2 = c2 . Notice also that (a, c) divides (c − a, c + a) = (2s2 , 2r2 ) = 2. But
since r ̸≡ s(mod 2), a and c are odd and so (a, c) = 1. Hence (a, b, c) = 1.
Suppose now that a, b, c is a pythagorean triplet. Write b2 = (c − a)(c + a). Since
a, c are both odd this implies (b/2)2 = ((a + c)/2)((c − a)/2). From (a, b, c) = 1
and a2 + b2 = c2 it follows that (a, c) = 1. Hence also ((c + a)/2, (c − a)/2) = 1.
Since the product of these numbers equals (b/2)2 each of them is a square, say
(c + a)/2 = r2 and (c − a)/2 = s2 . Hence c = r2 + s2 , a = r2 − s2 . Moreover,
(r, s) = 1 and r2 ≡ (c + a)/2 ≡ s2 + a ≡ s2 + 1(mod 2) ⇒ r2 ̸≡ s2 (mod 2) ⇒ r ̸≡
s(mod 2).
2
When we rewrite the equation a2 + b2 = c2 as (a/c)2 + (b/c)2 = 1 we see that
finding Pythagorean triplets is equivalent to finding rational numbers p, q such
that p2 + q 2 = 1, in other words, finding rational points on the unit circle.
Geometrically, the solution to this problem runs as follows. For any point (p, q) ∈
Q2 we draw the line between (p, q) and (1, 0) which is given by Y = t(1 − X),
where t = q/(1 − p). Conversely, any line through (1, 0) is given by Y = t(1 − X).
2 −1
The second point of intersection with the unit circle is given by tt2 +1
, t22t+1 . Thus
we can conclude: there exists a bijection between the sets
{t ∈ Q} and {x, y ∈ Q| x2 + y 2 = 1, (x, y) ̸= (1, 0)}
given by
y
t=
,
1−x
(
(x, y) =
t2 − 1 2t
,
t2 + 1 t2 + 1
)
.
Using Theorem 9.2.2 it is very simple to find all α, β, γ ∈ Q such that α2 +β 2 = γ 2 .
They are all of the form
( 2
)2 (
)2 ( 2
)2
r + s2
r − s2
2rs
=
+
, r, s, M ∈ Z.
M
M
M
A number n ∈ N is called congruent if n is the surface area of a right-angled
triangle with rational sides. In other words
n is congruent ⇐⇒ ∃r, s, M ∈ Z such that n =
1 2rs r2 − s2
.
2M M
F.Beukers, Elementary Number Theory
96
CHAPTER 9. DIOPHANTINE EQUATIONS
The latter equation is equivalent to M 2 n = rs(r2 − s2 ). It is a classical problem
to characterise congruent numbers. N.Koblitz used this topic in his book ‘Introduction to Elliptic Curves and Modular Forms’ as a leading motive. The smallest
congruent number is 5. A notorious congruent number is n = 157. The simplest
triangle corresponding to it has right angle sides
6803298487826435051217540
411340519227716149383203
411340519227716149383203
21666555693714761309610
and hypothenusa
224403517704336969924557513090674863160948472041
8912332268928859588025535178967163570016480830
(D.Zagier)
As an interesting curiosity we mention two results on congruent numbers
Theorem 9.2.3 (Birch, 1975) When n is prime and equal 5 or 7 modulo 8
then n is congruent. When n is twice a prime of the form −1(mod 4) then n is
also congruent.
Theorem 9.2.4 (Tunnell, 1983) Suppose n is a congruent number then the
number of integral solutions to 2x2 + y 2 + 8z 2 = n equals twice the number of
solutions to 2x2 + y 2 + 32z 2 = n.
It is generally expected that the converse of Tunnell’s theorem also holds.
9.3
Fermat’s equation
After reading about pythagorean triples it seems natural to ask the following
question. Let n ∈ N and n > 2. Does the equation
xn + y n = z n
(9.2)
have any solutions in x, y, z ∈ N? Fermat believed the answer to be ‘no’ and
claimed to have a ‘remarkable proof’. Unfortunately the margin of the book in
which he made this claim was ‘too narrow to write this proof down’. In June
1993 the English mathematician Andrew Wiles came quite close and for some
time it was believed that he did have a proof. However, his 200 page manuscript
of highly advanced mathematics turned out to have a gap and it took about a
year of suspense before this gap was repaired with the help of R.Taylor, a former
student of Wiles. This happened in October 1994. Wiles’s work not only resolves
Fermat’s last problem, it is also a major advance in the theory of elliptic curves,
in particular the Shimura-Taniyama-Weil conjecture.
Before Wiles’s discovery the equation (9.2) had been solved for certain special
values of n. For example, Fermat did prove the following theorem.
Theorem 9.3.1 The equation x4 + y 4 = z 2 has no solution x, y, z ∈ N.
F.Beukers, Elementary Number Theory
9.3. FERMAT’S EQUATION
97
As a consequence we see that (9.2) with n = 4 has no solutions.
Proof. Suppose there exists a solution. Let x0 , y0 , z0 be a solution with minimal
z0 . We may assume that (x0 , y0 ) = 1 and that y0 is even. We shall repeatedly
use Theorem 9.2.2. From x40 + y04 = z02 follows,
∃ r, s ∈ Z : (r, s) = 1, x20 = r2 − s2 , y02 = 2rs, z0 = r2 + s2 .
From x20 + s2 = r2 , x0 odd and (r, s) = 1 follows,
∃ ρ, σ ∈ Z : (ρ, σ) = 1, x0 = ρ2 − σ 2 , s = 2ρσ, r = ρ2 + σ 2 .
Together with y02 = 2rs this yields (y0 /2)2 = ρσ(ρ2 + σ 2 ). Since the factors
ρ, σ, ρ2 + σ 2 are pairwise relatively prime, and their product is a square, we get
∃ u, v, w ∈ Z : ρ = u2 , σ = v 2 , ρ2 + σ 2 = w2 .
After
elimination
of ρ, σ we get w2 = u4 + v 4 . A simple check shows |w| =
√
√
2
2
ρ + σ = r < y0 < z0 , contradicting the minimality of z0 . Hence there can
be no solutions.
2
The principle to construct a smaller solution out of a given (hypothetical) solution
is known as Fermat’s descending induction or descent . This principle, in disguised
form with cohomology groups and all, is still often used for many diophantine
equations.
The case n = 3 was settled by Euler (1753), Dirichlet dealt with the case n = 5
in 1820 and Lamé proved Fermat’s conjecture for n = 7 in 1839. Notice that
the case n = 6 follows from n = 3 because x6 + y 6 = z 6 can be rewritten as
(x2 )3 + (y 2 )3 = (z 2 )3 . In general, since any number larger than 2 is divible either
by 4 or by an odd prime, it suffices to prove Fermat’s conjecture for n = 4, which
we have already done, and for n prime. The methods of solution all follow the
same pattern. Let p be an odd prime and put ζ = e2πi/p . Then xp + y p = z p can
be rewritten as
(x + y)(x + ζy) · · · (x + ζ p−1 y) = z p .
The left hand side of the equation has been factored into linear factors at the
price of introducing numbers from Z[ζ]. The right hand side of the equation is a
p-th power and the principle of the proof is now to show that the linear factors
on the left are essentially p-th powers in Z[ζ]. To reach such a conclusion we
would need the property that factorisation into irreducible elements is unique in
Z[ζ]. Assuming this one would be able to conclude a proof of Fermat’s conjecture,
although it is still not easy. Unfortunately there is one more complicating factor,
prime factorisation in Z[ζ] need not be unique. Finding a way around this problem
has been one of the major stimuli to the development of algebraic number theory.
In 1847 E.Kummer proved the following remarkable theorem.
F.Beukers, Elementary Number Theory
98
CHAPTER 9. DIOPHANTINE EQUATIONS
Theorem 9.3.2 (Kummer) Denote by B0 , B1 , B2 . . . the sequence of Bernoulli
numbers. If the odd prime number p does not divide the numerators of B2 , B4 , B6 , . . . , Bp−3
then xp + y p = z p has no solution in positive integers.
Recall that the Bernoulli numbers B0 , B1 , B2 , . . . are given by the Taylor series
∑ Bn
x
=
xn .
ex − 1 n=0 n!
∞
It is not hard to see that Bn = 0 when n is odd and larger than 1. A small list
of values,
B2 = 1/6
B4 = −1/30
B6 = 1/42
B8 = −1/30
B10 = 5/66
B12 = −691/2730
B14 = 7/6
B16 = −3617/510
B18 = 43867/798
B20 = −174611/330
As an amusing aside we mention that the numerator of Bk with k ≤ p − 3, k
even, is divisible by p if and only if 1k + 2k + · · · + (p − 1)k is divisible by p2 .
Using the computer and further refinements of Kummer’s theorem one had been
able to verify Fermat’s conjecture for 2 < n < 4 · 106 (Buhler, Crandall, Sompolski) around 1990. For more details about the history and proof of Kummer’s
theory we refer to the books of P.Ribenboim (13 Lectures on Fermat’s last theorem, Springer Verlag 1979) and H.M.Edwards (Fermat’s last theorem, Springer
Verlag 1977). Of course these books were pre-Wiles. For an introduction for a
general audience to the techniques entering Wiles’s proof I highly recommend Simon Singh’s book Fermat’s Enigma: The epic quest to solve the world’s greatest
matehmatical problem (1998). It reads like a novel.
As a generalisation of Fermat’s conjecture Euler conjectured that for any k ∈
N there are no positive integers x1 , x2 , . . . , xk such that xk1 + · · · + xkk−1 = xkk .
However, this was disproved by a counterexample of Lander and Parkin (1967)
reading 1445 = 275 + 845 + 1105 + 1335 . Only in 2004 a second example was
discovered by J.Frye:
555 + 31835 + 289695 + 852825 = 853595 .
In 1988 N.Elkies found spectacular counterexamples in the case k = 4, the smallest of which reads 958004 + 2175194 + 4145604 = 4224814 . He also showed that
there exist infinitely many of such examples with k = 4.
9.4
Mordell’s equation
Let k ∈ Z. The equation
y 2 = x3 − k
F.Beukers, Elementary Number Theory
(9.3)
9.4. MORDELL’S EQUATION
99
in x, y ∈ Z is known as Mordell’s equation . This equation has been the subject of
many investigation by many people. In fact, a whole book has been written about
it (London, Finkelnstein: Mordell’s equation x3 − y 2 = k). The main theorem is,
Theorem 9.4.1 (Mordell) The eqation (9.3) has finitely many solutions.
The proof uses algebraic number theory and is beyond the scope of these notes.
Although Mordell’s theorem is a finiteness theorem, one cannot deduce an algorithm from it to actually determine the solutions of any given equation. Bounds,
which in principle give an effective solution became available around 1968 by
4
A.Baker who showed that log |x| ≤ c · |k|10 and slightly improved by H.Stark
log |x| ≤ Cϵ |k|1+ϵ for every ϵ > 0. The constants c, cϵ can be computed explicitly.
It turns out that the solution set depends in a very erratic way on the value of
k. For example a short computer search reveals the solutions
32
42
52
92
232
2822
3752
3786612
=
=
=
=
=
=
=
=
(−2)3 + 17
(−1)3 + 17
23 + 17
43 + 17
83 + 17
433 + 17
523 + 17
52343 + 17
It is a highly non-trivial task to show that this the complete solution set of
y 2 = x3 + 17. Two examples which are easier to deal with are given in the
following theorem.
Theorem 9.4.2 The equation y 2 = x3 + 7 has no solutions in x, y ∈ Z. The
only integral solutions to the equation y 2 = x3 − 2 are (x, y) = (3, ±5) (Fermat).
Proof. First we deal with y 2 = x3 + 7. Note that x is odd, because x even would
imply that y 2 ≡ 7(mod 8), which is impossible. Now notice that
y 2 + 1 = x3 + 8 = (x + 2)(x2 − 2x + 4)
Notice also that for any x, x2 −2x+4 = (x−1)2 +3 ≡ 3(mod 4). Hence x2 −2x+4
always contains a prime divisor p which is 3(mod 4). So we get y 2 +1 ≡ 0(mod p)
which is impossible because of p ≡ 3(mod 4).
√
To deal with y 2 = x3 − 2 we use arithmetic in the euclidean ring R = Z[ −2].
Notice first of all that x is odd. If x were even, then x3 − 2 ≡ 2(mod 4), so x3 − 2
cannot be a square. From y 2 + 2 = x3 follows the factorisation
√
√
(y + −2)(y − −2) = x3
F.Beukers, Elementary Number Theory
100
CHAPTER 9. DIOPHANTINE EQUATIONS
√
√
√
The gcd of y + −2 and y − 2√ −2 divides their difference which is 2 −2. So the
gcd is either 1 or divisible by −2. Since y is odd
holds. Thus
√ the first possibility
√
3
we find that there
+ −2 = (a + b −2) . Computing
√ exist3 a, b ∈2 Z such2 that 2y √
the√cube, y + −2 = a − 6ab + b(3a − 2b ) −2. Comparison of the coefficients
of −2 on both sides yields 1 = b(3a2 − 2b2 ). Hence b = ±1 and 3a2 − 2b2 = ±1.
So we find a = ±1 and b = ±1. Hence x = a2 +2b2 = 3. The values of y follow. 2
An interesting difference between the equations y 2 = x3 +7 and y 2 = x3 −2 is that
the second equation has infinitely many rational solutions. This can be seen by
the so-called chord and tangent method. In the point (3, 5) of the algebraic curve
y 2 = x3 −2 we draw the tangent to the curve. It is given by y−2 = (27/10)(x−3).
Now intersect this line with the curve y 2 = x3 − 2. Elimination of y yields
x3 −
729 2 837
1161
x +
x−
=0
100
50
100
Because of our tangent construction we already know that this equation has a
double root in x = 3. So the third root must also be a rational number. And
indeed we find
129
(x − 3)2 (x −
)=0
100
. So the x coordinate of the third intersection point of the tangent with the curve
equals 129/100. The corresponding y coordinate is 383/1000. Indeed we check
that (x, y) = (129/100, 383/1000) is a rational solution of y 2 = x3 − 2. Repetition
of this procedure provides us with an infinite set of rational solutions. In fact
it turns out that the rational points on y 2 = x3 − 2 together with the point ‘at
infinity’ have a group structure known as the Mordell-Weil group. This is the
beginning of a fascinating subject of rational points on elliptic curves. Excellent
introductions can be found in Silverman and Tate: Rational points on elliptic
curves.
By checking results for a large number of k M.Hall made the following conjecture
Conjecture 9.4.3 (Hall) There exists a constant C > 0 such that |x3 − y 2 | >
Cx1/2 for any x, y ∈ N with x3 − y 2 ̸= 0.
It is also known that there exist infinitely many positive integers x, y such that
√ 1/2
0 < |x3 − y 2 | < x (Danilov, 1982) so in this sense Hall’s conjecture is the
sharpest possible.
9.5
The ‘abc’-conjecture
In 1986 Masser and Oesterlé formulated a striking conjecture, the truth of which
has far reaching consequences for diophantine equations. For any a ∈ Z we let
N (a) (the conductor or radical of a) denote the product of all distinct primes of
a.
F.Beukers, Elementary Number Theory
9.5. THE ‘ABC’-CONJECTURE
101
Conjecture 9.5.1 (‘abc’ conjecture) Let ϵ > 0. Then there exists c(ϵ) > 0
such that for any triple of non-zero numbers a, b, c ∈ Z satisfying a + b + c = 0
and gcd(a, b, c) = 1 we have
max(|a|, |b|, |c|) < c(ϵ)N (abc)1+ϵ .
To get a feeling for what this conjecture says it is best to consider a number of
consequences.
Consequence 1. Let p, q, r be fixed numbers larger than 1 and (p, q, r) = 1.
Then
px + q y = r z
has only finitely many solutions x, y, z ∈ Z≥0 . Application of the conjecture
shows that
rz < c(ϵ)N (px q y rz )1+ϵ ≤ c(ϵ)N (pqr)1+ϵ .
Hence rz is a bounded number and so are px , q y . In particular x, y, z are bounded.
By other methods it is indeed possible to show that px + q y = rz has finitely
many solutions.
Consequence 2. Fermat’s conjecture is true for sufficiently large n. Apply the
‘abc’ conjecture to xn + y n = z n with x, y, z ∈ N to obtain
z n < c(ϵ)N (xn y n z n )1+ϵ ≤ c(ϵ)N (xyz)1+ϵ ≤ c(ϵ)z 3(1+ϵ) .
Hence, assuming z ≥ 2,
2n−3(1+ϵ) ≤ z n−3(1+ϵ) ≤ c(ϵ)
and this implies n ≤ log c(ϵ)/ log 2 + 3(1 + ϵ).
Consequence 3 Let p, q, r ∈ Z≥2 . Suppose
xp + y q = z r
has infinitely many solutions x, y, z ∈ N with gcd(x, y, z) = 1. Then
1 1 1
+ + ≥ 1.
p q r
Application of the ‘abc’ conjecture yields
z r ≤ c(ϵ)N (xp y q z r )1+ϵ
≤ c(ϵ)(xyz)1+ϵ
≤ c(ϵ)(z r/p z r/q z)1+ϵ .
Taking z → ∞ this implies r ≤ (1 + r/p + r/q)(1 + ϵ) for any ϵ > 0. Hence
r ≤ 1 + r/p + r/q and our assertion follows.
Considering the potential consequences of Conjecture 9.5.1 it is likely to be very
difficult to prove. In fact, any weaker version with 1 + ϵ replaced by another
number would already be spectacular! The best that can be done by present
day methods (1994) is max(|a|, |b|, |c|) < γ exp(N (abc)15 ) where γ is some (large)
constant.
F.Beukers, Elementary Number Theory
102
9.6
CHAPTER 9. DIOPHANTINE EQUATIONS
The equation xp + y q = z r
Motivated by consequence 3 of the previous section and the Fermat conjecture
we might look at the equation
xp + y q = z r
gcd(x, y, z) = 1,
xyz ̸= 0
where p, q, r are given integers > 1. From the ’abc’-conjecture we expect this
equation to have finitely many solutions when 1/p + 1/q + 1/r < 1. It was a very
pleasant surprise when Darmon and Granville actually proved this statement in
1993. It turns out to be a consequence of Mordell’s conjecture, now known as
Faltings’s theorem (1983), which we will describe in the next section. Until now
(1997) the only known solutions to this type of equation are
1k + 23 = 32 (k > 5)
132 + 73 = 29
27 + 173 = 712
25 + 72 = 34
35 + 114 = 1222
177 + 762713 = 210639282
14143 + 22134592 = 657
338 + 15490342 = 156133
438 + 962223 = 300429072
92623 + 153122832 = 1137 .
Notice that in each case there occurs an exponent 2. This leads us to the following
unsolved question, which can be seen as a generalisation of Fermat’s conjecture.
Question 9.6.1 Suppose p, q, r are integers ≥ 3. Do there exist solutions to
xp + y q = z r
gcd(x, y, z) = 1,
xyz ̸= 0 ?
When 1/p + 1/q + 1/r = 1 we can show that the set {p, q, r} equals one of the
sets {3, 3, 3}, {2, 4, 4} or {2, 3, 6}. The case p = q = r = 3 is known to have no
solution since Euler, The case p = q = 4, r = 2 was proved in these notes. In the
exercises we show that x4 + y 2 = z 4 has no non-trivial solutions. As for the case
{2, 3, 6} we only have the solutions 23 + (±1)6 = (±3)2 , but this is not easy to
prove.
When 1/p + 1/q + 1/r > 1 we can show that the set {p, q, r} equals one of the
sets {2, 2, k} (k ≥ 2), {2, 3, 3}, {2, 3, 4} or {2, 3, 5}. The case p = q = r = 2
corresponds of course to pythagorean triplets. We might wonder if the other
F.Beukers, Elementary Number Theory
9.6. THE EQUATION X P + Y Q = Z R
103
cases allow parametric solutions as well. In 1993 Don Zagier, amused by this
question, showed that the following parametrisation yield all integral solutions in
the cases {2, 3, 3} and {2, 3, 4}.
The equation x3 + y 3 = z 2 .
x = s4 + 6s2 t2 − 3t4
y = −s4 + 6s2 t2 + 3t4
z = 6st(s4 + 3t4 )
x = (1/4)(s4 + 6s2 t2 − 3t4 )
y = (1/4)(−s4 + 6s2 t2 + 3t4 )
z = (3/4)st(s4 + 3t4 )
x = s4 + 8st3
y = −4s3 t + 4t4
z = s6 − 20s3 t3 − 8t6
The equation x4 + y 3 = z 2 .
x = (s2 − 3t2 )(s4 + 18s2 t2 + 9t4 )
y = −(s4 + 2s2 t2 + 9t4 )(s4 − 30s2 t2 + 9t4 )
z = 4st(s2 + 3t2 )(s4 − 6s2 t2 + 81t4 )(3s4 − 2s2 t2 + 3t4 )
x = 6st(s4 + 12t4 )
y = s8 − 168s4 t4 + 144t8
z = (s4 − 12t4 )(s8 + 408s4 t4 + 144t8 )
x = 6st(3s4 + 4t4 )
y = 9s8 − 168s4 t4 + 16t8
z = (3s4 − 4t4 )(9s8 + 408s4 t4 + 16t8 )
x = s6 + 40s3 t3 − 32t6
F.Beukers, Elementary Number Theory
104
CHAPTER 9. DIOPHANTINE EQUATIONS
y = −8st(s3 − 16t3 )(s3 + 2t3 )
z = s12 − 176s9 t3 − 5632s3 t9 − 1024t12
x = −5s6 + 6s5 t + 15s4 t2 − 60s3 t3 + 45s2 t4 − 18st5 + 9t6
y = 6s8 − 56ts7 + 112t2 s6 − 168t3 s5 + 252t4 s4 − 168t5 s3 + 72t7 s − 18t8
z = −29s12 +156ts11 −726t2 s10 +2420t3 s9 −4059t4 s8 +3960t5 s7 −2772t6 s6 +
2376t7 s5 − 3267t8 s4 + 3564t9 s3 − 1782t10 s2 + 324t11 s + 27t12
x = s6 + 6s5 t − 15s4 t2 + 20s3 t3 + 15s2 t4 + 30st5 − 17t6
y = 2s8 − 8ts7 − 56t3 s5 − 28t4 s4 + 168t5 s3 − 112t6 s2 + 88t7 s + 42t8
z = −3s12 +12ts11 −66t2 s10 −44t3 s9 +99t4 s8 +792t5 s7 −924t6 s6 +2376t7 s5 −
1485t8 s4 − 1188t9 s3 + 2046t10 s2 − 156t11 s + 397t12
The equation x4 + y 2 = z 3 .
x = (s2 + 3t2 )(s4 − 18s2 t2 + 9t4 )
y = 4st(s2 − 3t2 )(s4 + 6s2 t2 + 81t4 )(3s4 + 2s2 t2 + 3t4 )
z = (s4 − 2s2 t2 + 9t4 )(s4 + 30s2 t2 + 9t4 )
x = 6st(s4 − 12t4 )
y = (s4 + 12t4 )(s8 − 408s4 t4 + 144t8 )
z = s8 + 168s4 t4 + 144t8
x = 6st(3s4 − 4t4 )
y = (3s4 + 4t4 )(9s8 − 408s4 t4 + 16t8 )
z = 9s8 + 168s4 t4 + 16t8
x = (3/2)st(s4 − 3t4 )
y = (1/8)(s4 + 3t4 )(s8 − 102s4 t4 + 9t8 )
z = (1/4)(s8 + 42s4 t4 + 9t8 )
F.Beukers, Elementary Number Theory
9.7. MORDELL’S CONJECTURE
105
We shall deal with the case {2, 2, k} in the exercises. The only equation that
remains is x5 + y 3 = z 2 . In 1995 it was proved by F.Beukers that again a finite
numbers of parametrised solutions suffice to give the complete solution set. In
2001 Johnny Edwards managed to produce the full list of these parametrisations.
Here is the x-coordinate of one such parametrisation,
x = 185s12 − 144s11 t − 2046s10 t2 + 9680s9 t3 − 13365s8 t4 + 15840s7 t5
−20724s6 t6 + 9504s5 t7 + 8415s4 t8 − 16720s3 t9 + 6930s2 t10 − 1776st11 + 701t12
9.7
Mordell’s conjecture
After seeing a good many particular examples one might wonder whether anything is known about diophantine equations in general. For a long time only one
result in such a direction was known.
Theorem 9.7.1 (C.L.Siegel 1929) Let P (X, Y ) ∈ Z[X, Y ] be a polynomial,
irreducible in C[X, Y ]. Suppose that the genus of the projective curve given by
P = 0 is at least 1. Then P (x, y) = 0 has at most finitely many solutions in
x, y ∈ Z.
The proof is quite difficult and involves ideas from diophantine approximation
and arithmetic algebraic geometry. Standard examples of curves of genus ≥ 2 are
the hyper-elliptic curve y 2 = q(x), where q(x) is a polynomial of degree at least
5 and distinct zeros and the Fermat curve xn + y n = 1 with n > 3.
Already in 1922 L.J.Mordell conjectured that under the conditions of Siegel’s
theorem P (x, y) = 0 has at most finitely many solutions in x, y ∈ Q. This
conjecture withstood attempts to solve it for a long time until in 1983 G.Faltings
managed to provide a proof of it. Unfortunately this proof can only be understood
by experts in arithmetic algebraic geometry. In 1988 P.Vojta found a brilliant
new proof which, unfortunately, had the same drawback as Faltings’ proof in
that it was accessable only to a very small group of experts. In 1990 E.Bombieri
considerably simplified Vojta’s proof, thus making it understandable for a large
audience of number theorists and algebraic geometers.
About polynomial diophantine equations in more than two variables almost nothing is known, although there exist a good many fascinating conjectures about
them.
9.8
Exercises
Exercise 9.8.1 Let a, b, c be any integral solution of a2 + b2 = c2 . Prove that 5
divides abc.
F.Beukers, Elementary Number Theory
106
CHAPTER 9. DIOPHANTINE EQUATIONS
Exercise 9.8.2 Solve the equation x2 + y 2 = z 4 in x, y, z ∈ N with gcd(x, y, z) =
1.
Exercise 9.8.3 Let a, b, c be any solution of a2 + b2 = c4 with gcd(a, b, c) = 1.
Prove that 7 divides abc.
Exercise 9.8.4 Solve the equation x2 + y 2 = z 3 in x, y, z ∈ N with gcd(x, y, z) =
1. (Hint: use factorisation in Z[i]).
Exercise 9.8.5 (*) Prove that x4 − y 4 = z 2 has no solution x, y, z ∈ N.
Exercise 9.8.6 Show that 2k − 3l = ±1, k, l ≥ 2 has only the solution k =
3, l = 2. (The conjecture that xk − y l = 1, k, l ≥ 2 has only xk = 32 , y l = 23 as
solution has long been known as Catalan’s conjecture. In 2002 it was proven by
Michailescu).
Exercise 9.8.7 Prove that there exist no rectangular triangles with integral edges
whose surface area is a square.
Exercise 9.8.8 Solve 4y 2 = x3 + 1 in x, y ∈ Z.
Exercise 9.8.9 Prove that the only integer solution of
x2 + y 2 + z 2 = 2xyz
is x = y = z = 0.
Exercise 9.8.10 Prove that 32 −1 is divisible by 2n+2 . Construct triples ak , bk , ck ∈
N for k = 1, 2, . . . such that ak +bk = ck , gcd(ak , bk , ck ) = 1 and limk→∞ ck /N (ak bk ck ) =
∞. Here N (x) is the product of the distinct prime divisors of x.
n
Exercise 9.8.11 Suppose the abc-conjecture holds. Prove that there exist at most
finitely many triples ar , bs , ct such that ar + bs = ct , gcd(a, b) = 1 and 1/r + 1/s +
1/t < 1. (Hint: prove and make use of the following statement: 1/r + 1/s + 1/t <
1 ⇒ 1/r + 1/s + 1/t ≤ 1 − 1/42.)
Exercise 9.8.12 Find a, b, c ∈ Z such that 7 ̸ |abc and a7 + b7 ≡ c7 (mod 73 ).
Exercise 9.8.13 Show, assuming the abc-conjecture, the modified Hall conjecture which reads as follows. For every a < 1/2 there exists c(a) > 0 such that for
any positive integers x, y with x3 ̸= y 2 we have |x3 − y 2 | > c(a)xa .
F.Beukers, Elementary Number Theory
Chapter 10
Prime numbers
10.1
Introductory remarks
In previous chapters we have seen that prime numbers play a crucial role in
number theory. We repeat here that by a prime number we mean an integer
which cannot be written as a product of smaller numbers. In this chapter we
occupy ourselves with the distribution of primes in N.
Theorem 10.1.1 (Euclid) There exist infinitely many primes.
Proof. We have already seen Euclid’s proof in Theorem 1.4.5. Here we present
another proof due to Euler. Although slightly more complicated than Euclid’s
proof it uses an idea which will return repeatedly.
We shall show that
)
∏(
1
1−
p
p≤N
tends to zero as N → ∞. Here, the product is taken over all primes p ≤ N .
Notice that by the unique factorisation theorem in Z,
)−1 ∏ (
) ∑
N
∏(
1
1
1
1
1−
=
1 + + 2 + ··· >
p
p p
n
n=1
p≤N
p≤N
Since the latter sum tends to infinity as N → ∞ we see that our product tends
to zero as N → ∞. Hence there are infinitely many primes.
2
As a consequence of Euler’s method we find the following corollary.
∑1
Corollary 10.1.2 The sum
, taken over all primes, diverges.
p
1
Proof. Notice that for any x ∈ (0, 12 ) we have x > 12 log 1−x
. Hence
∏
∑1
1
1
> log
p
2
1 − p−1
p≤N
p≤N
107
108
CHAPTER 10. PRIME NUMBERS
and since the product tends to ∞ as N → ∞ we see that our sum diverges.
2
A more precise analysis, as performed by Mertens, reveals that there exists a real
number A such that for all X > 2 we have
∑1
= log log X + A + O((log X)−1 )
p
p<X
where the summation is over all primes p < X
Definition 10.1.3
π(x) = #{p ≤ x| p prime}.
The local distribution of prime numbers seems to be completely erratic, and not
much is known about it. Remarkably enough one can say quite a few things about
the global distribution of primes as reflected by π(x). In the following table we
have counted s, the number of primes in the interval [x − 75000, x + 75000] for
several values of x,
x
s
150000/ log x
8
10
8154
8143
109
7242
7238
10
10
6511
6514
11
10
5974
5922
1012
5433
5428
13
10
5065
5011
1014
4643
4653
15
10
4251
4342
The last column of this table suggests that the density of the primes near x is
about equal to 1/ log x. This led Gauss to conjecture,
∫ x
dt
π(x) ∼ li(x) :=
.
2 log t
The sign ∼ must be interpreted as asymptotic equality. More precisely, f (x) ∼
g(x) means that f (x)/g(x) → 1 as x → ∞. Since li(x) ∼ x/ log x we might also
conjecture,
x
π(x) ∼
.
log x
The first result in this direction was obtained by Chebyshev who proved around
1852 that
x
x
< π(x) < 1.11
0.92
log x
log x
for sufficiently large x. In 1860 Riemann, in a now historical paper, introduced
the complex function
∑ 1
ζ(s) =
Res > 1
ns
n≥1
F.Beukers, Elementary Number Theory
10.1. INTRODUCTORY REMARKS
109
in the study of prime numbers. For Res > 1 it is easy to see that ζ(s) is analytic
and Riemann showed that it can be continued analytically to C with the exception
of s = 1, where it has a first order pole with residue 1. The relation of ζ(s) with
prime numbers is made apparent by the Euler factorisation
ζ(s) =
∏(
p
1
1− s
p
)−1
where the product is over all primes p. It turns out that for the distribution
of the primes the zeros of ζ(s) in the critical strip 0 ≤ Res ≤ 1 are extremely
important. If there would have been no zeros in this strip one could have proved
some marvelous theorems on prime numbers. Unfortunately there are zeros in the
strip but, and this is fortunate again, they all seem to lie on the line Res = 1/2.
This was posed by Riemann as a question and until now no one has been able to
confirm it. This question, known as the Riemann hypothesis , is one of the classical
problems in mathematics. It follows very easily from Riemann’s work that the
zeros lie symmetrical around the X-axis. So it suffices to look at zeros in the
upper half plane. We order them in the order of increasing imaginary part. Then
it is known that the first 1500 000 000 zeros are all simple and lie on Res = 1/2
(Brent, v.d.Lune, te Riele, Winter). It was proved by Levinson (1974) that at
least a third of the zeros in the critical strip is on the line Res = 1/2. It should
be noted that the zeros outside the critical strip are given by s = −2, −4, −6, . . .
and they are called the trivial zeros.
Continuing Riemann’s work Hadamard and De la Vallée-Poussin, independently
of each other, proved the following theorem in 1896.
Theorem 10.1.4 (Prime number theorem)
π(x) ∼
x
.
log x
In 1949 Selberg and Erdös, more or less independently, gave an elementary proof
of the prime number theorem. By elementary we mean that no use is made of
complex function theory. It does not imply that the proof is simple!
Later it turned out that Gauss’ function li(x) is a better approximation of π(x)
than x/ log x. Assuming the Riemann hypothesis we have
π(x) = li(x) + O(x1/2 log x).
However, one is a long way off at proving such results. The best known estimate
are of the form
1√
π(x) = li(x) + O(x exp(−
log x).
15
F.Beukers, Elementary Number Theory
110
CHAPTER 10. PRIME NUMBERS
Here we give a table which compares values of π(x) and li(x) for several x.
x
π(x)
li(x) − π(x)
102
25
5
3
10
168
10
4
10
1229
17
105
9592
38
6
10
78498
130
107
664579
339
8
10
5761455
754
9
10
50847534
1701
1010
455052511
3104
11
10
4118054813
11588
1012
37607912018
38263
13
10
346065536839
108971
14
10
3204941750802
314890
1015 29844570422669
1052619
1016 279238341033925
3214632
As a peculiarity note that the values of δ(x) = li(x) − π(x) in our table are
all positive. Since x becomes quite large in our table one might suspect that
δ(x) is always positive. That one should be careful in stating such beliefs was
shown by Littlewood who proved in 1914 that δ(x) changes sign infinitely often.
Around 1933 Skewes showed that, assuming the Riemann hypothesis, the first
1034
sign change should be somewhere below 1010 . Numbers of this size were soon
called Skewes’ numbers. We now know, without having to assume the Riemann
hypothesis, that the first change of δ(x) is somewhere below 6.7 × 10370 .
Finally a few words about the local distribution of primes. Let us denote by
p1 , p 2 , . . . , pn , . . .
the sequence of prime numbers in increasing order. First of all, pn+1 − pn may
become arbitrarily large, i.e. there exist arbitrarily long gaps in the sequence
of prime numbers. To find a gap of length at least N − 1, say, we just have
to write down N ! + 2, N ! + 3, . . . , N ! + N and notice that these numbers are
divisible by 2, 3, . . . , N . By means of elementary methods Chebyshev proved
in 1852 that pn+1 < 2pn , thereby confirming Bertrand’s postulate . It is now
1
− 384
known, using deep analytic methods that pn+1 − pn = O(pθn ) with θ = 11
12
(Iwaniec, Pintz, Mozzochi). If Riemann’s hypothesis is true then one can prove
√
that pn+1 − pn = O( pn ). However, based on heuristic arguments, one expects
more, namely pn+1 − pn = O((log pn )2 ) (Cramér).
If pn+1 = pn +2 then we call the pair pn , pn+1 a twin prime . Examples are (11, 13)
(29, 31) (41, 43) (71, 73) . . .. It is not known whether there exist infinitely many
F.Beukers, Elementary Number Theory
10.2. ELEMENTARY METHODS
111
twin primes, although one generally expects the answer to be ‘yes’.
∑ The first nontrivial result was by V.Brun in 1919 who showed that the series p−1 , taken over
all primes belonging to a twin prime, converges. This was the first example of
an important tool in analytic number theory, namely sieve methods. The largest
known twin prime in 2001 (September) was
318032361 × 2107001 ± 1
but such records usually have a short life.
For much more information we refer to P.Ribenboim’s delightful ‘The book of
prime number records’.
10.2
Elementary methods
In this section we shall prove two theorems by elementary methods which are
similar in spirit to the methods used by Chebyshev.
Theorem 10.2.1 Let n ∈ N and n > 10. Then
1 n
n
< π(n) < 3
.
3 log n
log n
Theorem 10.2.2 (Bertrand’s postulate) For any n ∈ N there exists a prime
number p such that n < p ≤ 2n.
For the proof of these theorems we require a few lemmas.
Lemma 10.2.3 Let n ∈ N. Then
a) ∀n ≥ 5 :
(
b) ∀n ≥ 4 :
(
Proof. Notice that
(
)
2n
< 4n−1
n
)
2n
4n
> .
n
n
)
(
)
2n
2n(2n − 1) 2(n − 1)
=
.
n
n·n
n−1
Since 4(n − 1)/n < 2n(2n − 1)/(n · n) < 4 this implies
(
) ( )
(
)
n − 1 2(n − 1)
2n
2(n − 1)
4
<
<4
.
n
n−1
n
n−1
(10.1)
To prove statement (a) notice that it is true for n = 5 and apply induction on n
using the second inequality in (10.1). To prove (b) note that it is true for n = 4
and apply induction on n using the first inequality in (10.1).
2
F.Beukers, Elementary Number Theory
112
CHAPTER 10. PRIME NUMBERS
Lemma
( ) 10.2.4 Let p be prime and n ∈ N. Let k be the number of prime factors
p in 2n
. Then
n
(a) pk ≤ 2n.
(b) If n < p ≤ 2n then k = 1.
(c) If 2n/3 < p ≤ n and n > 2 then k = 0.
Proof. For any m ∈ N we know that the number of factors p in m! is equal to
[
] [ ]
]
∞ [
∑
m
m
m
+ 2 + ··· =
.
p
p
pr
r=1
Hence
k=
]
∞ ([
∑
2n
pr
r=1
[
n
−2 r
p
])
.
(10.2)
Notice that 0 ≤ [2x] − 2[x] ≤ 1 for any x ∈ R. Notice also that the terms in
(10.2) vanish when r > log 2n/ log p. Hence
k≤
[
∑
r≤log 2n/ log p
]
log 2n
log 2n
1=
≤
log p
log p
and assertion (a) follows immediately.
If n < p ≤ 2n all terms in (10.2) vanish except [ 2n
], which is 1. This implies (b).
p
Suppose 2n/3 < p ≤ n. Statement (c) can be verified by hand for 2 < n < 5. So
we can assume n ≥ 5. Then we have p2 > 2n and all terms in (10.2) with r > 1
vanish. Hence k = [2n/p] − 2[n/p]. But 2n/3 < p ≤ n implies 1 ≤ n/p < 3/2
and thus we find that k = 0, as asserted in (c).
2
Lemma 10.2.5 For any n ≥ 2 we have
∏
p < 4n .
p≤n
pprime
( )
Proof. Let m ∈ N and m ≥ 5. Observe that according to Lemma 10.2.4(b) 2m
m
is divisible by all primes p with m < p ≤ 2m. Hence, using Lemma 10.2.3(a),
∏
p < 4m−1 .
m<p≤2m
F.Beukers, Elementary Number Theory
(10.3)
10.2. ELEMENTARY METHODS
113
We shall prove our statement by induction on n. First of all our lemma can be
verified by∏hand for∏all n < 10. Now suppose n ≥ 10. If n is even we have
obviously p≤n p = p≤n−1 p < 4n−1 < 4n and we are done. If n is odd we write
∏
∏
∏
p=
p
p.
p≤ n+1
2
p≤n
n+1
<p≤n+1
2
The first product on the right can be estimated using our induction hypothesis
∏
and the second by using (10.3) with m = (n + 1)/2. We obtain p≤n p <
4(n+1)/2 · 4(n+1)/2−1 = 4n , as desired.
2
Proof of Thm 10.2.1. First we prove the upper bound. Choose α ∈ R between
0 and 1. Notice that Lemma 10.2.5 implies that
∏
p < 4n .
√
n<p≤n
Since each factor in the product is larger than
√
(1/2)(π(n)−π( n))
n
√
n this implies that
< 4n
and hence
√
n
π(n) − π( n) < 2 log 4
.
log n
√
√
If we estimate π( n) by n/2 we find that
π(n) < 2.8
n
n0.5
+
.
log n
2
When n > 200 this can be bounded by 3n/ log n as desired. For n ≤ 200 we have
to verify this upper bound case by case.
Let m ≥ 4. By combination of Lemma 10.2.3(b) and Lemma 10.2.4(a) the lower
bound is derived as follows
( )
∏
4m
2m
≤
<
(2m) = (2m)π(2m) .
m
m
p≤2m
Hence
log(4m /m)
2m
> log 2
− 1.
log 2m
log 2m
Suppose n ≥ 8. When n = 2m is even we deduce,
n
π(n) ≥ log 2
−1
log n
π(2m) ≥
and when n = 2m + 1 is odd we have
π(n) = π(2m + 1) = π(2m + 2) > log 2
2m + 2
n
− 1 > log 2
− 1.
log(2m + 2)
log n
When n ≥ 10 the bound log 2 logn n −1 can be bounded below by
1 n
3 log n
as desired. 2
F.Beukers, Elementary Number Theory
114
CHAPTER 10. PRIME NUMBERS
Proof of Thm 10.2.2. Using Lemma 10.2.4(a)(b)(c) we see that
( )
∏
∏
∏
2n
p
p
<
(2n).
n
√
n<p≤2n
p≤2n/3
(10.4)
p≤ 2n
The second product on the right can be estimated using Lemma 10.2.5 and yields
the upper
bound 42n/3 . The third product on the right can be estimated by
√
2n
(2n) . Elementary estimates show that this can be bounded above by 4n/3 /n
as soon as n > 512. Hence when n > 512 we obtain from (10.4),
( ) ( ∏ ) n
2n
4
<
p
.
(10.5)
n
n
n<p≤2n
( )
> 4n /n. Together with
From Lemma 10.2.3(b) we have the∏lower bound 2n
n
(10.5) this implies that the product n<p≤2n p is non-empty when n > 512. For
n ≤ 512 observe that
2, 3, 5, 7, 13, 23, 43, 83, 113, 223, 443, 881
is a sequence of prime numbers of which each term is smaller than twice its
predecessor. Hence the theorem is also true for n ≤ 512.
2
10.3
Exercises
Exercise 10.3.1 Prove,
a)
∑
∏
(1 − p−s )−1 ≥
n−s
b)
∏
−s −1
(1 − p )
<
∞
∑
n−s
∏
(1 − p−s )−1 =
∞
∑
n−s
∀s > 1.
n=1
p prime
Exercise 10.3.2 Prove,
a)
∏
∑
(1 − p−s )−1 ≥
n−s
p≤X
p prime
∀s > 1, ∀X ≥ 1
n=1
p≤X
p prime
c)
∀s ∈ R>0 , ∀X ≥ 1
n≤X
p≤X
p prime
n≤X
F.Beukers, Elementary Number Theory
∀s ∈ R>0 , ∀X ≥ 1
10.3. EXERCISES
115
b)
∑1
> log X
n
n≤X
∀X ≥ 1.
Now choose s = 1 in part a), take log’s on both sides and show that
c)
∑ 1
1
> log log X − .
p
2
p≤X
p prime
Exercise 10.3.3 a) Prove that
(
)
∫ x
dt
x
=O
k
(log x)k
2 (log t)
∀k ∈ N, x ≥ 2.
(Hint: split the integration interval into two parts.)
b) Prove that
x
x
x
+ 1!
+ · · · + (k − 1)!
+O
li(x) =
2
log x
(log x)
(log x)k
(
x
(log x)k+1
)
∀n ∈ N.
Exercise 10.3.4 Prove that
∏ ( p2 + 1 ) 5
= .
p2 − 1
2
p prime
(Hint: use ζ(2) = π 2 /6 and ζ(4) = π 4 /90).
Exercise 10.3.5 Let p1 , p2 , p3 , . . . be the sequence of consecutive prime numbers.
Prove, using the prime number theorem, that pn /n log n → 1 as n → ∞.
Exercise 10.3.6 Let ϵ > 0. Prove, using the prime number theorem, that there
exists x0 (ϵ) such that for any x > x0 the interval [x, (1 + ϵ)x] contains a prime
number. (The case ϵ = 1 is known as Bertrand’s postulate.)
Exercise 10.3.7 Verify, using the prime number theorem, whether or not the
following sums converge,
∑
p
1
,
p
log
p
prime
∑ log p
.
p
p prime
Exercise 10.3.8 Prove, using the prime number theorem, that
log(lcm(1, . . . , n))
= 1.
n→∞
n
lim
F.Beukers, Elementary Number Theory
116
CHAPTER 10. PRIME NUMBERS
Exercise 10.3.9 (*) (H.W.Lenstra jr.) Prove that for infinitely many n ∈ N we
have π(n)|n. (Examples: π(30) = 10|30, π(1008) = 168|1008).
∫1
Exercise 10.3.10 Consider for any n the integral In = t=0 tn (1 − t)n dt.
a)Prove that In is a rational number whose denominator divides lcm[n, . . . , 2n+1].
(Hint: integrate term by term).
b)Prove that |In | < (1/4)n .
c)Prove that lcm[1, . . . , m] ≥ 2m−1 for all m ∈ N.
Exercise 10.3.11 Use the ideas of the
√ previous exercise.
a) Prove that |t(1 − t)(1 − 2t)| < 1/6 3, ∀t ∈ [0, 1].
b) Prove that lcm(1, 2, . . . , 6n + 1) ≥ 108n ∀n ∈ N.
1
c) Prove that lcm(1, 2, . . . , m) ≥ 108
((108)1/6 )m , ∀m ∈ N.
F.Beukers, Elementary Number Theory
Chapter 11
Irrationality and transcendence
11.1
Irrationality
Definition 11.1.1 Let α ∈ C. We call α irrational when α ̸∈ Q.
Proving irrationality and transcendence of numbers is now being considered as a
branch of number theory, although the techniques that are used involve subjects
like complex analysis, linear differential equations and algebraic geometry. In this
chapter we shall restrict ourselves to examples in which only elementary methods
are used.
The easiest numbers for which irrationality can be proved are the algebraic numbers.
Theorem 11.1.2 Let α be a zero of a polynomial xm + c1 xm−1 + · · · + cm ∈ Z[x].
Then α is either irrational or α ∈ Z. In the latter case we have α|cm .
Proof. Suppose xm + c1 xm−1 + · · · + cm has a rational zero p/q with p, q ∈
Z, (p, q) = 1, q > 0. Then pm + c1 pm−1 q + · · · + cm q m = 0. Hence q|pm and since
(p, q) = 1 this implies q = 1. Notice that pm + c1 pm−1 + · · · + cm = 0 now implies
that p|cm .
2
Corollary√11.1.3 Let m ∈ N. If N ∈ Z is not the m-th power of an integer
then α = m N is irrational.
Proof. Notice that αm − N = 0. If α ∈ Q then α ∈ Z according to Theorem 11.1.2 and hence N is the m-th power of an integer. This contradicts our
assumptions, hence α ̸∈ Q.
2
Theorem 11.1.4 Let e be the base of the natural logarithm. Then e is irrational.
117
118
CHAPTER 11. IRRATIONALITY AND TRANSCENDENCE
Proof. Suppose e is rational with denominator d. We use the series expansion
e=
∞
∑
1
.
k!
k=0
The number
1
1
1
− − ··· −
1! 2!
k!
is a positive rational number with a denominator dividing d(k!). Hence, since α
is not zero,
1
α≥
.
k!d
On the other hand,
α := e − 1 −
1
1
+
+ ···
(k + 1)! (k + 2)!
(
)
1
1
1
+
+ ···
=
k! k + 1 (k + 1)(k + 2)
(
)
1
1
1
<
+
+ ···
k! k + 1 (k + 1)2
11
=
k! k
α =
contradicting our lower bound for α whenever k > d.
2
The irrationality proof of π is more complicated and we require the following
lemma.
Lemma 11.1.5 Let m ∈ Z≥0 . Then
∫ 1
π
tm sin πtdt
0
is a polynomial in 1/π 2 with integral coefficients and degree [m/2].
Proof. By induction on m. For m = 0 and m = 1 we have
∫ 1
∫ 1
π
sin πtdt = 2
π
t sin πtdt = 1.
0
0
Suppose m > 1. After a two-fold partial integration we obtain
∫ 1
∫ 1
m(m − 1)
m
π
tm−2 sin πtdt
π
t sin πtdt = 1 −
2
π
0
0
from which our assertion follows.
F.Beukers, Elementary Number Theory
2
11.1. IRRATIONALITY
119
Theorem 11.1.6 We have π 2 ̸∈ Q and π ̸∈ Q.
Proof. Define Pm (t) =
the integral
1 d m m
( ) t (1−t)m
m! dt
∫
and notice that Pm (t) ∈ Z[t]. Consider
1
In = π
(sin πt)P2n (t)dt.
0
After a 2n-fold partial integration we find
π 2n
In = π(−1)
(2n)!
∫
1
(sin πt)t2n (1 − t)2n dt.
n
0
Hence
π 2n+1
0 < |In | <
.
(2n)!
On the other hand we can compute In term by term. Lemma 11.1.5 then implies
that In = An (1/π 2 ), where An (x) ∈ Z[x] and deg An ≤ n.
Suppose that π 2 = a/b for some a, b ∈ N. Because An has degree ≤ n the number
In = An (1/π 2 ) = An (b/a) is a fraction whose denominator divides an . Moreover,
In ̸= 0. Hence
1
|In | ≥ n .
a
Together with the upper bound for |In | this implies
1
π 2n+1
<
a2n
(2n)!
which becomes impossible when n → ∞. Hence π 2 ̸∈ Q. This immediately
implies π ̸∈ Q.
2
Around 1740 Euler proved e to be irrational and the first proof of the irrationality
of π was given by Lambert in 1761. This proof was based on the continued fraction
expansion of arctg(x). The proof we gave can be considered as a variation of a
proof given by I.Niven. On the other hand there are many ’naturally’ occurring
numbers for which no irrationality results are
∑known. For example, it is not
known whether Euler’s constant γ = limn→∞ ( nk=1 (1/k) − log n) or e + π or eπ
is irrational. Motivated by the standard series for e P.Erdös asked the following
question. Is
∞
∑
1
k! + 1
k=1
irrational? Surprisingly this seems to be difficult to answer.
F.Beukers, Elementary Number Theory
120
11.2
CHAPTER 11. IRRATIONALITY AND TRANSCENDENCE
Transcendence
Definition 11.2.1 A number α ∈ C is called algebraic if it is the zero of a nontrivial polynomial with coefficients in Q. A number is called transcendental if it
is not algebraic.
Obviously, proving transcendence of a number is much harder than proving irrationality. It is therefore no surprise that in the beginning of the 19th century no
examples of transcendental numbers were known. In 1844 Liouville proved the
following theorem.
Theorem 11.2.2 (Liouville 1844) Let α be an algebraic number whose minimal polynomial has degree n. Then there exists a positive constant c such that
for any p, q ∈ Z and q > 0,
p
α − ≥ c .
q qn
Proof. Let P (x) ∈ Z[x] be the minimal polynomial of α. Since P has no rational
roots we have P (p/q) ̸= 0 and, moreover, |P (p/q)| ≥ 1/q n . Write P (x) =
(x − α)Q(x) and let M = max|x−α|≤1 |Q(x)|. Suppose |α − p/q| ≤ 1. Then,
trivially, |P (p/q)| ≤ M |α−p/q|. Combined with our lower bound for |P (p/q)| this
yields |α − p/q| ≥ 1/(M q n ). This proves our theorem with c = min(1, 1/M ). 2
Corollary 11.2.3 Let α ∈ R. Suppose that there exists a sequence of rational
numbers {pn /qn }∞
n=1 , with qn > 0 for all n, and a sequence of numbers λn such
that
pn c
0 < α − < λn
lim λn = ∞
n→∞
qn
q
n
for some c > 0. Then α is transcendental.
This corollary enabled Liouville to construct infinitely many examples of transcendental numbers. We give one example here,leaving the construction of other
examples to the reader.
Corollary 11.2.4 The number
∞
∑
1
α=
2k!
k=0
is transcendental.
∑
Proof. We apply Corollary 11.2.3. Let qn = 2n! and pn = 2n! nk=0 (1/2k! ). Then
∞
∞
∑ 1 p
1 ∑ 1
2
2
n
α − = = (n+1)! = n+1 .
< (n+1)!
k!
j
qn
2 2
2
2
qn
j=0
k=n+1
F.Beukers, Elementary Number Theory
11.2. TRANSCENDENCE
121
Application of Corollary 11.2.3 yields our result.
2
Through the pioneering work of Cantor on set theory around 1874 it also became
clear that ‘almost all’ real numbers are transcendental. This follows from the
following two theorems.
Theorem 11.2.5 The set of algebraic numbers is countable.
Proof. It suffices to show that the set Z[X] is countable. To any polynomial
P (X) = pn X n + pn−1 X n−1 + · · · + p1 X + p0 ∈ Z[X] with pn ̸= 0 we assign the
number µ(P ) = n + |pn | + |pn−1 | + · · · + |p0 | ∈ N. Clearly for any N ∈ N the
number of solutions to µ(P ) = N is finite, because both the degree and the size
of the coefficients are bounded by N . Hence Z[X] is countable.
2
Theorem 11.2.6 (Cantor) The set of real numbers is uncountable.
Proof. We will show that the set of real numbers in the interval [0, 1) is uncountable. Suppose that this set is countable. Choose an enumeration and
denote the decimal expansion of the n-th real number by 0.an1 an2 an3 · · ·, where
anm ∈ {0, 1, 2, 3, 4, 5, 6, 7, 8, 9} for all n, m. Now consider the real number β whose
decimal expansion reads 0.b1 b2 b3 · · · where the bi are chosen such that bi ̸= aii for
every i. This choice implies that β does not occur in our enumeration. Hence
[0, 1) is uncountable.
2
The principle of the proof of Theorem 11.2.6 is known as Cantor’s diagonal procedure and it occurs in many places in mathematics.
Almost all real numbers being transcendental, it seems ironic that until the end
of the 19-th century not a single ‘naturally occurring’ number was known to be
transcendental. Only in 1873 Hermite showed that e is transcendental and in
1882 Lindemann proved π to be transcendental. In his famous lecture of 1900
D.Hilbert asked whether numbers of the form ab with √a, b algebraic, a ̸= 0, 1
and b ̸∈ Q, are transcendental. Specific examples are 2 2 and i−2i = eπ . This
problem was considered to be very difficult by Hilbert, but already in the 1930’s
A.O.Gel’fond and Th.Schneider indepently developed techniques to solve this
problem. So now we know,
Theorem 11.2.7 (Gel’fond, Schneider ,1934) Let a, b be algebraic and suppose that a ̸= 0, 1 and b ̸∈ Q. Then ab is transcendental.
Corollary 11.2.8 Let α, β be two positive real algebraic numbers such that β ̸= 1
and log α/ log β ̸∈ Q. Then log α/ log β is transcendental.
Proof. Let b = log α/ log β and suppose b is algebraic. Then, according to
Theorem 9.2.7 the number α = β b is transcental which is impossible since α is
algebraic.
2
Nowadays the Gel’fond-Schneider theory has grown into a field of its own in which
large classes of numbers, ususally related to algebraic geometry, are known to be
transcendental.
F.Beukers, Elementary Number Theory
122
CHAPTER 11. IRRATIONALITY AND TRANSCENDENCE
11.3
Irrationality of ζ(3)
∑
−s
Let ζ(s) = ∞
n=1 n . We shall be interested in the numbers ζ(m) with m ∈ Z≥2 .
Euler showed that
B2n 2n
ζ(2n) = (−1)n−1 22n−1
π
(2n)!
where B2n is the 2n-th Bernoulli number. Hence ζ(2n) is transcendental because
π 2n is transcendental. Strangely enough next to nothing is known about the
numbers ζ(2n + 1) for n ≥ 1. It was therefore a complete surprise when in
1978 the french mathematician R.Apéry announced a proof of ζ(3) ̸∈ Q. The
first reaction of his fellow mathematicians was incredulity, since the presentation
of the proof was a mixture of remarkable formulae and downright impossible
statements. Later this proof was patched up by H.Cohen and D.Zagier and
Apéry turned out to be correct on all the crucial parts. The simple proof we
present here was found by F.Beukers, but the shape of the integrals is motivated
by Apéry’s formulae.
Lemma 11.3.1 Let r, s ∈ Z≥0 . If r > s then
∫
1
∫
1
−
0
0
Z
log xy r s
x y dxdy ∈
.
1 − xy
[1, 2, . . . , r]3
If r = s then
∫
0
1
∫
0
1
(
)
log xy r r
1
1
−
x y dxdy = 2 ζ(3) − 3 − · · · − 3 .
1 − xy
1
r
Proof. Let σ > 0 and consider the integral
∫
0
1
∫
0
1
xr+σ y s+σ
dxdy.
1 − xy
Develop (1 − xy)−1 in a geometrical series and carry out the integration term by
term. We find
∞
∑
1
.
(k + r + σ + 1)(k + s + σ + 1)
k=0
When r > s this implies
∫
0
1
∫
0
1
(
)
∞
∑
xr+σ y s+σ
1
1
1
dxdy =
−
1 − xy
r−s k+s+σ+1 k+r+σ+1
k=0
(
)
1
1
1
+ ··· +
=
r−s s+1+σ
r+σ
F.Beukers, Elementary Number Theory
11.3. IRRATIONALITY OF ζ(3)
123
Differentiate with respect to σ and put σ = 0,
(
)
∫ 1∫ 1
log xy r s
1
1
1
Z
x y dxdy = −
+ ··· + 2 ∈
.
2
r − s (s + 1)
r
[1, 2, . . . , r]3
0
0 1 − xy
When r = s we find in a similar way,
∫
1
∫
0
0
1
)
(
∞
∑
log xy r r
1
−2
1
x y dxdy =
= −2 ζ(3) − 3 − 3
3
1 − xy
(k
+
r
+
1)
1
r
k=0
2
Lemma 11.3.2 When n is sufficiently large, lcm[1, 2, . . . , n] < 3n .
Proof. Notice that
lcm[1, 2, . . . , n] =
∏
p[log n/ log p] <
p≤n
∏
plog n/ log p ≤
p≤n
∏
n = nπ(n)
p≤n
where the products are taken over the primes p. According to the prime number
theorem we have π(n) < (log 3)n/ log n for sufficiently large n. Hence nπ(n) < 3n
for n sufficiently large.
2
Theorem 11.3.3 (R.Apéry 1978) The number ζ(3) is irrational.
Proof. Consider the double integral
∫ 1∫ 1
− log xy
In =
Pn (x)Pn (y)dxdy
1 − xy
0
0
( d )n n
where Pn (x) = n!1 dx
x (1 − x)n . Notice that P (x) ∈ Z[x]. From Lemma 11.3.1
it follows that
An + Bn ζ(3)
In =
,
An , Bn ∈ Z.
[1, 2, . . . , n]3
Notice that
− log xy
=
1 − xy
∫
hence
In =
∫
0
1
1
dz
1 − (1 − xy)z
Pn (x)Pn (y)
dxdydz
1 − (1 − xy)z
∫
∫1∫1∫1
where stands for 0 0 0 . After an n-fold partial integration with respect to
x we obtain
∫
(xyz)n (1 − x)n Pn (y)
In =
dxdydz.
(1 − (1 − xy)z)n+1
F.Beukers, Elementary Number Theory
124
CHAPTER 11. IRRATIONALITY AND TRANSCENDENCE
Substitute z = (1 − (1 − xy)w)/(1 − w). Then
∫
Pn (y)
dxdydw.
In = (1 − x)n (1 − w)n
1 − (1 − xy)w
An n-fold partial integration with respect to y yields
∫ n
x (1 − x)n y n (1 − y)n wn (1 − w)n
In =
dxdydw.
(1 − (1 − xy)w)n+1
It is an excercise to show that for all 0 ≤ x, y, w ≤ 1 we have
√
x(1 − x)y(1 − y)w(1 − w) ≤ ( 2 − 1)4 .
1 − (1 − xy)w
Hence
∫
√
|In | < ( 2 − 1)
4n
√
dxdydw
= 2( 2 − 1)4n ζ(3).
1 − (1 − xy)w
On the other hand, In ̸= 0 and In = (An + Bn ζ(3))/[1, 2, . . . , n]3 . Suppose
ζ(3) = p/q with p, q ∈ Z and q > 0. Then, using Lemma 11.3.2,
√
1
1
<
≤ |In | < 2( 2 − 1)4n ζ(3).
n
3
27 q
[1, 2, . . . , n] q
Hence,
√
1 < ( 2 − 1)4n 27n 2ζ(3)q.
√
Since ( 2 − 1)4n 27n tends to 0 as n → ∞, we have a contradiction. Therefore,
ζ(3) ̸∈ Q.
2
11.4
Exercises
Exercise 11.4.1 Prove, using the series expansions for e and e−1 , that e is not
algebraic of degree 2.
Exercise 11.4.2 Show that
log 3
log 2
is irrational.
Exercise 11.4.3 Prove that e + π and eπ are not both rational.
Exercise 11.4.4 Prove that
∞ ( )n
∑
4
1
̸∈ Q.
5
3n2
n=0
Exercise 11.4.5 Show that the so-called Champernowne number
0.1234567891011121314151617 . . .
is irrational (Actually K.Mahler proved it transcendental in the 1930’s, but that
is much harder).
F.Beukers, Elementary Number Theory
Chapter 12
Solutions to selected problems
(1.5.3) When m is odd we have the polynomial factorisation
xm + 1 = (x + 1)(xm−1 − xm−2 + · · · − x + 1).
Suppose that n is not a power of 2, i.e. n contains an odd divisor m > 1.
Suppose n = m · k. Substitute x = 2k in the above identity, then
2n + 1 = 2mk + 1 = (2k + 1)(2k(m−1) − 2k(m−2) + · · · − 2k + 1).
So 2k +1 is a divisor of 2n +1. It remains to point out that it is a non-trivial
divisor, i.e. 1 < 2k + 1 < 2n + 1. So 2n + 1 is not prime, a contradiction.
Therefore n cannot contain odd divisors > 1.
(1.5.2) When m is odd we have the polynomial factorisation
xm − 1 = (x − 1)(xm−1 + xm−2 + · · · + x + 1).
Suppose that n is not a prime i.e. n = m · k for some integers k, m > 1.
Substitute x = 2k in the above identity, then
2n − 1 = 2mk − 1 = (2k − 1)(2k(m−1) + 2k(m−2) + · · · + 2k + 1).
So 2k −1 is a divisor of 2n −1. It remains to point out that it is a non-trivial
divisor, i.e. 1 < 2k − 1 < 2n − 1. So 2n − 1 is not prime, a contradiction.
Therefore n cannot be composite.
(1.5.6) Suppose that p + 2 is composite for only finitely many primes p. Then there
is a number P0 such that p prime and p > P0 implies p + 2 prime. Choose
a prime p > P0 . Then, consequently, all odd numbers larger than p are
prime. This is clearly impossible since, fore example, all powers of 3 are
composite. We get a contradiction.
125
126
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
(1.5.8) Since n is odd we can write n = 2m + 1. Hence
(
)
m+1
n = (2m + 1) = 4m + 4m + 1 = 8
+ 1.
2
2
2
2
So we see that n2 is 1 modulo 8.
(1.5.12) For a),b) we refer to the course notes. Let eta = (1 +
ratio. Note that η 2 = η + 1.
√
5)/2, the golden
Part c) When n = k we must prove that rk ≥ 1 which is clearly true, since
rk is the last non-zero remainder. When n = k − 1 we must show that
rk−1 ≥ η which is true as well, since rk1 ≥ 2. Now we apply induction using
rn−2 =
≥
≥
=
qn−1 rn−1 + rn
rn−1 + rn
η k−n+1 + η k−n = η k−n (η + 1)
η k−n η 2 = η k−n+2
Part d) We have a = r−1 and r−1 ≥ η k+1 . Hence η k+1 ≤ a, from which our
desired inequality follows.
(1.5.14) We give a hint. Suppose n > m and suppose n = mq + r with 0 ≤ r < m.
Notice that we have the polynomial identity
(xn − 1) = (xm − 1)(xn−m + xn−2m + · · · + xn−qm ) + xr − 1.
By application of the euclidean algorithm to an − 1 and am − 1 we see
that the exponents are precisely the remainders of the euclidean algorithm
applied to n, m.
(1.5.16) Let k be the number of zeros. In other words, k is the highest power such
that 10k divides 123!. Since there are more factors 2 than 5 in 123!, it
suffices to count the number of factors 5. Between 1 and 123 there are
24 five-tuples and 4 twentyfive-tuples (twentyfive-tuples are also counted
as five-tuples). There are no hundredtwentyfive-tuples or higher between 1
and 123. In total this gives us 24+4 factors 5. Hence k = 28.
(1.5.17(a)) Consider the numbers 1 to n. Among these numbers there are [n/p] ptuples, [n/p2 ] p2 -tuples, [n/p3 ] p3 -tuples, etc. Here we count pk -tuples also
as pk−1 -tuples. The total number of factors p in n! therefore equals
[ ] [ ] [ ]
n
n
n
+ 2 + 3 + ···
p
p
p
F.Beukers, Elementary Number Theory
127
(2.3.1) A perfect number n is characterised by σ(n) = 2n. In other words,
2n divide on both sides by n to obtain
∑
d|n
d=
∑d
= 2.
n
d|n
Notice that as d runs over the divisors of n, the number d/n runs over all
inverses of the divisors of n. Hence
∑1
= 2.
d
d|n
(2.3.3) All convolution products are convolution products of multiplicative functions. Hence the convolution products are also multiplicative. To describe
the products it suffices to describe their values in the prime powers. Here
are the results,
1. (Ik ∗ Ik )(n) = σ0 (n)nk
2. µ ∗ I1 = ϕ
3. (µ ∗ µ)(pk ) = −2 als k = 1, 1 als k = 2, 0 als k > 2
4. (µ ∗ 2ω )(n) = |µ(n)|
5. (2ω ∗ 2ω )(pk ) = 4k
6. (µ ∗ ϕ)(pk ) = pk − 2pk−1 + pk−2 if k ≥ 2, p − 2 if k = 1.
(2.3.7) We shall prove something more general. Let f (n) be any arithmetic function
with f (1) ̸= 0. Then there exists an arithmetic function g such that f ∗ g =
e. In other words, we must determine numbers g(1), g(2), g(3), . . . such that
f (1)g(1) = 1 and for all n > 1,
∑
f (d)g(n/d) = 0
d|n
Hence g(1) = 1/f (1) and for n = 2, 3, 4, . . .,
∑
f (1)g(n) = −
g(d)f (n/d).
d|n,d<n
Note that the latter relation allows us to determine g(2), g(3), g(4), . . . recursively. Notice also that g is uniquely determined.
To complete our exercise, we must show that if f is multiplicative, then
so is g. To that end we construct a multiplicative function h such that
f ∗ h(pk ) = 0 for all prime powers pk and f ∗ h(1) = 1. By the multiplicative
F.Beukers, Elementary Number Theory
128
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
property of f, h it follows that f ∗ h = e and by the uniqueness of g we
conclude that g = h. For the construction of h we use the analogue of the
construction for g, but now restricted to prime powers pk ,
f (1)h(p ) = h(p ) = −
k
k
k−1
∑
h(pl )f (pk−l ).
l=0
The values of h at other integers are now defined by the requirement of
multiplicativity.
(2.3.8) Notice that both lefthand side and righthand side are multiplicative functions. Thus it suffices to show equality when n is a prime power pk . Note
that
k
k
∑
∑
∑
σ0 (d)3 =
σ0 (pl )3 =
(l + 1)3 .
d|pk
l=0
l=0
Also note that
k
k
∑
∑
∑
l 2
2
σ0 (p )) = ( (l + 1))2 .
(
σ0 (d)) = (
d|pk
l=0
l=0
The two results are equal because we have the famous identity
13 + 23 + · · · + (k + 1)3 = (1 + 2 + · · · + (k + 1))2
for all positive integers k.
(3.5.1) Suppose that n = ab with 1 < a < b < n. Then the product (n − 1)!
contains both factors a and b. Hence ab divides (n − 1)!. Suppose n is
composite and suppose it cannot be written as a product of two distinct
numbers a, b < n. Then n must be the square of a prime p, i.e. n = p2 .
We have assumed n > 4, so p > 2. But in that case the product (n − 1)!
contains the factors p and 2p. So p2 divides (n − 1)!.
(3.5.3) Note that a number is divisible by 11 if and only if the alternating sum of its
digits is divisible by 11. The alternating sum of the digits of a palindromic
number of even length is zero.
(3.5.4) Answers: 3(mod 7), 13(mod 71), 83(mod 183).
(3.5.6) Answers:
a) x ≡ 1307(mod 2100)
b) y ≡ 675(mod 1540)
c) z ≡ 193(mod 420)
F.Beukers, Elementary Number Theory
129
(3.5.8) Instead of a million consecutive numbers we can more generally ask for n
consecutive numbers, where n is any integer. We choose n distinct primes
p1 , p2 , . . . , pn and solve the simultaneous system of congruences
x + 1 ≡ 0(mod p21 ) x + 2 ≡ 0(mod p22 )
···
x + n ≡ 0(mod p2n ).
Since the numbers p2i are pairwise relatively prime, the Chinese remainder
theorem shows the existence of a solution x ∈ N. Hence x+1, x+2, . . . , x+n
is a sequence of n consecutive numbers all divisible by a square > 1.
(3.5.9) Notice that by the chinese remainder theorem,
a2 ≡ a(mod 10k ) ⇐⇒ a2 ≡ a(mod 2k ),
a2 ≡ a(mod 5k ).
Suppose we want to solve a2 ≡ a(mod pk ) for any prime p. Observe that
the equation is equivalent to pk |a2 − a, hence pk |a(a − 1). Since a and
a − 1 are relatively prime, we have either pk |a or pk |a − 1. In other words,
a ≡ 0(mod pk ) or a ≡ 1(mod pk ). Applying this to p = 2, 5 we get the
following possibilities
1. a ≡ 0(mod 2k ), a ≡ 0(mod 5k ). But this implies that a is divisible by
10k . This gives trivial solutions which are ruled out by the constraint
1 < a < 10k .
2. a ≡ 1(mod 2k ), a ≡ 1(mod 5k ). In this case a − 1 is divisible by 10k ,
which is another trivial solution ruled out by the extra requirement
1 < a < 10k .
3. a ≡ 0(mod 2k ), a ≡ 1(mod 5k ). According to the chinese remainder
theorem this has a unique residue class modulo 10k as solution. It is
not 0 or 1 modulo 10k , hence there exists a unique solution a with
1 < a < 10k .
4. a ≡ 1(mod 2k ), a ≡ 0(mod 5k ). Again this gives a unique solution.
In all we find two solutions to our general problem.
Now let k = 12 according to the above we must first solve a ≡ 0(mod 212 ), a ≡
1(mod 512 ), which gives us a ≡ 81787109376(mod 1012 ) and hence a =
81787109376. Secondly we must solve a ≡ 1(mod 212 ), a ≡ 0(mod 512 ),
which gives us a = 918212890625. These are the two solutions.
Notice that the sum of the non-trivial solutions found above, add up to
1000000000001. Can you explain that?
(3.5.12) Use the map a(mod 10) 7→ (a(mod 2), a(mod 5) to find
0(mod 10) 7→ (0(mod 2), 0(mod 5))
F.Beukers, Elementary Number Theory
130
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
1(mod
2(mod
3(mod
4(mod
5(mod
6(mod
7(mod
8(mod
9(mod
10)
10)
10)
10)
10)
10)
10)
10)
10)
7
→
7→
7
→
7
→
7
→
7
→
7
→
7
→
7
→
(1(mod
(0(mod
(1(mod
(0(mod
(1(mod
(0(mod
(1(mod
(0(mod
(1(mod
2), 1(mod
2), 2(mod
2), 3(mod
2), 4(mod
2), 0(mod
2), 1(mod
2), 2(mod
2), 3(mod
2), 4(mod
5))
5))
5))
5))
5))
5))
5))
5))
5))
(3.5.13) Case a) We must show that 42
≡ 3(mod 13). Notice that the order of
4(mod 13) equals 6. This means that in the determination of 4k (mod 13)
for any k, only the value of k(mod 6) matters. So we must determine
22n+1 (mod 6) for all n. Note that 22n+1 ≡ 0(mod 2) and 22n+1 ≡ (−1)2n+1 ≡
2n+1
−1(mod 3) for all n. Hence 22n+1 ≡ 2(mod 6). We conclude that 42
≡
42 ≡ 3(mod 13).
2n+1
Case b) When 37 divides x, the statement is certainly true. Suppose now
that 37 does not divide x. In that case, x36 ≡ 1(mod 37). So we must
determine 99 (mod 36). Notice that 99 ≡ 0(mod 9) and 99 ≡ 19 ≡ 1(mod 4).
Hence 99 ≡ 9(mod 36). A quick calculation shows that x9 (mod 37) has in
total 4 different values, 1, −1, 6, −6. After adding 4 to these numbers we
get 5, 3, 10, −2(mod 37) None of these numbers is 0(mod 37).
(3.5.14) By Euler’s theorem we have
ki
)
aϕ(pi
ki −1
(pi −1)
≡ api
≡ 1(mod pki i ) for i = 1, 2, . . . , r
Hence
aλ(n) ≡ 1(mod pki i ) for i = 1, 2, . . . , r
By the Chinese remainder theorem this implies aλ(n) ≡ 1(mod n).
(3.5.15) Notice that 2730 = 2 × 3 × 5 × 7 × 13. By the Chinese remainder theorem
it suffices to show that n13 ≡ n(mod p) for all n and p = 2, 3, 5, 7, 13. We
repeatedly use Fermat’ little theorem.
n13 ≡ n(mod 13) for all n (Fermat’s little theorem).
n13 ≡ n7 · n6 ≡ n · n6 ≡ n7 ≡ n
(mod 7).
n13 ≡ (n5 )2 · n3 ≡ n2 · n3 ≡ n5 ≡ n(mod 5).
n13 ≡ (n3 )4 · n ≡ n4 · n ≡ n3 · n · n ≡ n3 ≡ n(mod 3).
n13 ≡ n(mod 2) because n is even ⇐⇒ n13 is even.
F.Beukers, Elementary Number Theory
131
(3.5.16) We must solve: 2p−1 − 1 = pu2 in an odd prime p and an integer u. Note
that u must be odd. factorisation of the left hand side yields (2(p−1)/2 −
1)(2(p−1)/2 + 1) = pu2 . The factors on the left are relatively prime (check!).
This implies that there exist positive integers r, s such that either
2(p−1)/2 − 1 = pr2 ,
2(p−1)/2 + 1 = s2
or
2(p−1)/2 − 1 = r2 ,
2(p−1)/2 + 1 = ps2 .
In the first case s2 −1 is a power of 2. So s−1 = 2k and s+1 = 2l for certain
integers k, l met l > k > 0. taking the difference, 2l − 2k = 2. From this
follows that 2k |2 and hence k = 1. Consequently, s = 3 and 2(p−1)/2 +1 = 8.
This gives us p = 7. A small check, (26 − 1)/7 = 9, a square.
In the second case we see that r2 + 1 is a power of 2. Since r is odd, we
have r2 ≡ 1(mod 4) and hence r2 + 1 ≡ 2(mod 4). In other words, r2 + 1
contains at most one factor 2. So, r2 +1 = 2 ⇒ r = 1 ⇒ 2(p−1)/2 −1 = 1.We
conclude that p = 3. A small check, (22 − 1)/3 = 1, a square.
(3.5.17) Wilson’s theorem tells us that (p − 1)! ≡ −1(mod p). We now rewrite the
product (p − 1)! as
(
)
p−1
p+1
(p − 1)! =
!×
× · · · × (p − 1).
2
2
The second group of factors on the right is modulo p equal to the factors
−(p − 1)/2, −(p − 3)/2, . . . , −2, −1. Hence
((
(p − 1)! = (−1)
(p−1)/2
) )2
p−1
! .
2
If we now notice that (−1)(p−1)/2 = 1 (because p ≡ 1(mod 4)) and (p−1)! ≡
−1(mod p), our assertion follows.
(3.5.20) Notice that an ≡ 1(mod an − 1). Furthermore, ak ̸≡ 1(mod an − 1) for all
0 < k < n because 1 < ak − 1 < an − 1 for all such k. Hence the order of
a(mod an − 1) is precisely n. The order of an element divides the order of
the group, hence n ÷ ϕ(an − 1).
∏
(3.5.21) This is slightly tedious. Write ϕ(n) = n p|n (1 − 1/p). We give a lower
∏
bound for p|n (1 − 1/p). Notice that
∏
p|n
(1 − 1/p) ≥
1
2
∏
(1 − 1/p)
p|n, p odd
F.Beukers, Elementary Number Theory
132
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
The number of distinct primes dividing n can be at most log2 (n) (the logarithm of n in base 2). Note that for each odd prime p we have 1−1/p ≥ 2/3.
Hence
( )log2 (n)
∏
1 2
(1 − 1/p) ≥
2 3
p|n
1 − log(2/3)/ log(2)
n
2
1 −0.5
≥
n
2
=
We conclude that ϕ(n) ≥
n → ∞.
1
2
· n1−0.5 =
1
2
· n0.5 . The latter goes to ∞ as
(3.5.28) The orders are 6, 11, 8 respectively. To shortcut the computation, notice
for example that ϕ(46) = 22. So the order of 3(mod 46) divides 22. Thus
we need only check whether, 31 , 32 , 311 are 1(mod 46). If not, then 22 is the
order of 3(mod 46). It turns out that 311 ≡ 1(mod 46).
(3.5.30) Note that 2p ≡ 1(mod q). The order of 2(mod q) thus divides p. Since
p is a prime and 21 ̸≡ 1(mod q) the order is precisely p. Hence p divides
ϕ(q) = q − 1. So q ≡ 1(mod p). Furthermore, since q is odd, q ≡ 1(mod 2).
Hence, because p is odd, we conclude that q ≡ 1(mod 2p). So q is of the
form q = 2mp + 1.
(3.5.31) To show part (a) notice that a2 ≡ −1(mod q). Together with its square
n+1
a2
≡ 1(mod q) we note that a has order 2n+1 in (Z/qZ)∗ . Hence 2n+1
divides q − 1 which solves part (a).
n
We now follow a variation on Euclids proof. Suppose there are finitely
many primes p with p ≡ 1(mod 2n ). Call them p1 , p2 , . . . , pr . Let q be a
n
prime divisor of N = (2p1 p2 · · · pr )2 + 1, which is necessarily odd. Then
it follows from (a) that q ≡ 1(mod 2n ). So there is an i such that q = pi .
Hence N − 1 is divisible by q. Together with q|N this gives q|1 which is
impossible. We have a contradiction. There are infinitely many primes of
the form k · 2n + 1.
(3.5.33) Trial and error shows that 2 is a primitive root modulo 11. All other primitive roots modulo 11 can then be obtained by computation of 2k (mod 11) for
1 ≤ k < 10 and gcd(k, 10) = 1. So, 21 ≡ 2(mod 11), 23 ≡ 8(mod 11), 27 ≡
7(mod 11), 29 ≡ 6(mod 11) are the primitive roots modulo 11.
From the theory it follows that 2(mod 112 ) has order 10 or 110. Checking
that 210 ≡ 56(mod 121) we conclude that 2(mod 121) has order 110 and
hence is a primitive root. All other primitive roots modulo 121 can be
F.Beukers, Elementary Number Theory
133
computed by computing 2k (mod 121) with 1 ≤ k < 110 and gcd(k, 110) =
1.
(3.5.34)
1. By testing 2d (mod 13) for all divisors d of 12 we quickly see that 2 is
a primitive root modulo 13. The other primitive roots are given by
2k (mod 13), where gcd(k, 12) = 1. So, 2, 6, 7, 11(mod 13).
Notice (Z/14Z)∗ ≃ (Z/7Z)∗ × (Z/2Z)∗ ≃ (Z/7Z)∗ . The latter group is
cyclic, so there is a primitive root modulo 14. Note ϕ(14) = 6. A quick
check shows that 3 is a primitive root modulo 14. The other primitive
roots are given by 3k (mod 14), where gcd(k, 6) = 1. So, 3, 5(mod 14).
Since x4 ≡ 1(mod 5) and x2 ≡ 1(mod 3) for all x with gcd(x, 15) = 1,
we see that x4 ≡ 1(mod 15) for all such x. But ϕ(15) = 8, so there
cannot be primitive roots modulo 15.
2. Note that 5 is relatively prime with 12, the order of (Z/13Z)∗ . Notice
that 5 · 5 ≡ 1(mod 12). To solve x5 ≡ 7(mod 13) raise both sides to
the power 5. We find, (x5 )5 ≡ x25 ≡ x ≡ 75 ≡ 11(mod 13). The
solution is x ≡ 11(mod 13).
In a similar way we get x5 ≡ 11(mod 14) ⇒ x ≡ 9(mod 14) and
x5 ≡ 2(mod 15) ⇒ x ≡ 2(mod 15).
(3.5.35) We use the function λ(n) from exercise 3.5.14. If we have a primitive root
modulo m then we should have λ(m) = ϕ(m). In other words
lcm(pk11 (p1 − 1), . . . pkr r (pr − 1)) = (pk11 (p1 − 1), · · · pkr r (pr − 1)).
This means that the numbers pki i (pi − 1) are all relatively prime. In particular, there can be at most one odd prime factor pi . So m is of the form
m = 2l pk . When k = 0 we have m = 2l and we know that l = 1, 2. When
l = 0 we have m = pk . Suppose that k, l > 0 Then 2l−1 and pk−1 (p − 1) are
relative prime. But this is impossible if l > 1. Hence l = 1 and m = 2pl .
(3.5.29) Case a) We must determine all p such that 10 has order 1,2,3,4,5 or 6 modulo 10. Hence we must determine all p that divide at least one of 10−1, 102 −
1, 103 −1, 104 −1, 105 −1, 106 −1. This gives us p = 3, 7, 11, 13, 37, 41, 101, 271.
Case b) Using the idea from case a) we get p = 239, 4649.
Case c) p = 73, 137
(3.5.41) From an−1 ≡ 1(mod n) we see that gcd(a, n) = 1. Let k be the order of
a(mod n). Then k|n − 1. Suppose that (n −1)/k contains a prime divisor q.
Then a(n−1)/q ≡ 1(mod n), contradicting our assumptions. Hence (n − 1)/k
contaisn no prime divisors, hence (n−1)/k = 1 from which we get k = n−1.
We also know that k and hence n − 1 divide ϕ(n). This is only possible if
ϕ(n) = n − 1, hence n is prime.
F.Beukers, Elementary Number Theory
134
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
(3.5.42) Let N = 3 · 28 + 1. The prime divisors of N − 1 are 2, 3. Notice that
11(N −1)/2 ≡ −1(mod N ) and 11(N −1)/2 ≡ 360(mod N ). Hence, by Lehmer’s
test, N is prime.
(5.7.1) Simply write down all squares 12 , 22 , . . . , 82 modulo 17. We get 1, 4, 9, 16, 8, 2, 15, 13(mod 17)
as quadratic residues modulo 17. Similarly we get 1, 4, 9, 16, 6, 17, 11, 7, 5(mod 19)
as quadratic residues modulo 19.
(5.7.2) We first show that p does not divide b. If it did, then p|a2 + b2 and p|b
would imply p|a, which contradicts gcd(a, b) = 1. Hence p does not divide
b.
Now it follows from p|a2 + b2 that a2 ≡ −b2 (mod p). Multiply on both sides
by b−2 (mod p). We get (ab−1 )2 ≡ −1(mod p). Hence −1 is a quadratic
residue modulo p which implies that p ≡ −1(mod 4).
(5.7.4) Part a) Let a be a quadratic nonresidue modulo p and let x = a(p−1)/8 .
Then, x4 ≡ a(p−1)/2 ≡ −1(mod p).
Part b) Notice that x4 ≡ −1(mod p) implies that x2 ≡ −x−2 (mod p) and
hence x2 + x−2 ≡ 0(mod p). So we find that (x + 1/x)2 ≡ x2 + 2 + x−2 ≡
0(mod p). In other words, (x + 1/x)2 ≡ 2(mod p) and 2 is a quadratic
residue modulo p.
2
(5.7.7) We treat two examples. First
) ≡ 114(mod 127). Note that 127 is prime,
( 114 x
so it suffices to determine 127 . Notice,
(
114
127
)
(
=
2
127
)(
3
127
)(
19
127
)
The first factor is 1(because
127
)
( 1 ) ≡ −1(mod 8). The second factor, by
127
reciprocity equals (− )3 = −
third
( 13 )3 = −1.
( 19 The
)
( 6 )factor, again by reci127
procity,
equals
−
=
−
=
−
=
−
. The
19 ( )
13
13
( 2 )( 3 )
(191 )
( 114 )latter equals
13
− 13 13 = −(−1) 3 = 3 = 1. So we conclude 127 = −1, our
equation is not solvable.
We now study the solvability of 9x2 + 12x + 15 ≡ 0(mod 58). Splitting off
squares gives (3x + 2)2 + 11 ≡ 0(mod 58). So it suffices to study solvability
of y 2 ≡ −11(mod 58). By the Chinese remainder theorem this is equivalent
to the system y 2 ≡ 1(mod 2), y 2( ≡ )−11(mod 29). The first equation is
. Note that is equals
solvable, it remains to determine −11
29
(
−1
29
)(
11
29
)
(
=
29
11
)
(
=
Hence there are no solutions.
F.Beukers, Elementary Number Theory
7
11
)
(
=−
11
7
)
( )
4
=−
= −1.
7
135
(5.7.8) Let p be an odd prime ̸= 5. Note that
( ) ( )
5
p
=
p
5
The latter is 1 if p ≡ ±1(mod 5) and −1 if p ≡ ±2(mod 5).
Now let p be an odd prime ̸= 3. Then
( ) ( )( )
( )
−1
3
−3
(p−1)/2
(p−1)/2 p
=
= (−1)
(−1)
.
p
p
p
3
The latter is 1 if p ≡ 1(mod 3) and −1 if p ≡ −1(mod 3).
Finally,
( )
( )
3
(p−1)/2 p
= (−1)
.
p
3
we can now read off that the Legendre symbol is 1 if p ≡ ±1(mod 12) and
−1 if p ≡ ±5(mod 12).
(5.7.9) Let a = (−1)k0 p1 · · · pr be the prime factorisation of a where k0 ≡ 0 or 1
and the primes pi are not necessarily distinct. Note that the primes pi(are
)
odd because a ≡ ±1(mod 4). We now compute the Legendre symbol ap
using quadratic reciprocity.
( )k 0 ∏
( )
r ( )
−1
pi
a
=
p
p
p
i=1
r
r ( )
∏
∏
p
k0 (p−1)/2
(pi −1)(p−1)/4
= (−1)
(−1)
pi
i=1
i=1
Now notice the identity
k0 +
r
∑
(pi − 1)/2 ≡ (a − 1)/2(mod 2)
i=1
On the left we simply have modulo 2 the number of primes pi which are
≡ −1(mod 4) and the sign of a. When this total is odd we have a ≡
−1(mod 4), when it is even we have a ≡ 1(mod 4). So we get
( )
r ( )
∏
a
p
(a−1)(p−1)/2
.
= (−1)
p
p
i
i=1
∏( p )
only depends on the residue class p(mod |a|),
( )
the value of the sign depends on the parity of (p−1)/2. Hence ap depends
The value of the product
pi
F.Beukers, Elementary Number Theory
136
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
only on the residue class p(mod 4|a|). Moreover,
( ) if a ≡ 1(mod 4) the sign
in front of the product is always +, so now ap depends only on the class
p(mod |a|).
( )
(5.7.10) Note that x2 ≡ a(p+1)/2 ≡ a(p−1)/2 a ≡ ap a ≡ a(mod p).
(5.7.13) Let a be the smallest quadratic nonresidue and let b the smallest positive
number such that ab > p. Then, 0 < ab − p < a. Hence, by minimality of
a, the residue class ab(mod p) must be quadratic residue class. Hence b is
a quadratic non-residue. We also have ab < p + a, hence b < p/a + 1. Since
√
also a + 1 ≤ b it follows that a < p/a and so, a < p.
√
√
(5.7.14(a)) Notice that the sum [ p] + · · · + [ kp] equals the number of lattice point,
√
with positive coordinates, below the graph of px in the interval 1 ≤ x ≤ k.
√
√
√
Notice that [ pk] = [ p(p − 1)/4] = (p − 1)/2. The graph of px does
not contain lattice points when 1 ≤ x ≤ k. To do the counting we might as
well take the number of lattice points in the rectangle 1 ≤ x ≤ k, 1 ≤ y ≤
(p − 1)/2 and the subtract the number of lattice points on the right of the
graph. Hence,
(p−1)/2 [ 2 ]
∑ y
k(p − 1)
−
.
2
p
l=1
Note that
[
]
{ 2}
y2
y
y2
=
−
.
p
p
p
Take the sum for y∑= 1, 2, . . . , (p − 1)/2. The first part can be summed
using the formula nl=1 n2 = 16 n(n + 1)(2n + 1). The sum of the second
part is precisely equal to K/p where K is the sum of the quadratic residues
modulo p. In case p ≡ 1(mod 4) this equals half the sum of all residue
classes, which is p(p − 1)/2.
∑
(7.5.1) The sum n≤X r2 (n) equals the number of lattice points within the disc
√
with radius X. To every lattice point (a, b) we associate the elementary
square {(a + x, b + y)|0 ≤ x, y ≤ 1}. Let S be the union of these squares.
The number R(X) is precisely equal
the surface area of S. Not that S
√ to √
lies within
√ the√circle with radius X + 2. Note also that the circle with
radius X − 2 is entirely contained in S. Hence
√
√
√
√
π( X − 2)2 ≤ R(X) ≤ ( X + 2)2 .
√
from which we can derive that |R(X) − πX| ≤ 2π 2X + 2π.
(7.5.4) Part a),b) are done simply by trying. Part c) can be done by trying, but
also follows from the next exercise.
F.Beukers, Elementary Number Theory
137
(7.5.5) We write n = 2k [(3/2)k ]−1 as sum of k-th powers. Since n < 3k , it can only
be written as repeated sum with terms 1k , 2k . The most economic way to
do this is to use as many terms 2k as possible. The remainder can then be
written as a sum of ones. Note that [n/2k ] = [(3/2)k ] − 1 and the remainder
after division of n by 2k is 2k − 1. So we require [n/2k ] − 1 terms 2k and
2k − 1 terms 1k . Hence g(k) ≥ 2k + [(3/2)]k − 2.
(7.5.6) From the identity it follows that a number of the form 6m2 can be written
as a sum of 12 fourth powers. We simply write n = a2 +b2 +c2 +d2 (possible
by Lagrange’s theorem) and use the identity.
From the hint: n = 6N + r and write N as sum of four squares, we deduce
that 6N can be written as the sum of 4 × 12 = 48 squares. Now choose r
such that 0 ≤ r ≤ 5 and write r as sum of r terms 14 . We conclude that
g(4) ≤ 48 + 5 = 53.
To get a refinement, note that r need not be chosen between 0 and 5. We
can also choose among the remainders 0, 1, 2, 33, 16, 17, each which is the
sum of at most two fourth powers. Hence g(4) ≤ 48 + 2 = 50.
(9.8.1) We can assume that a, b, c form a Pythagorean triple. Suppose, without loss
of generality, that b is even. Then there exist r, s such that a = r2 − s2 , b =
2rs, c = r2 + s2 . Hence abc = 2rs(r4 − s4 ). If r or s is divisible by 5
we are done. If r, s are not divisible by 5 we have r4 ≡ 1(mod 5) and
s4 ≡ 1(mod 5). Hence r4 − s4 ≡ 1 − 1 ≡ 0(mod 5). So r4 − s4 is divisible
by 5.
(9.8.2) Note that x, y, z 2 is a Pythagorean triple. Assume that y is even, the
case x even being similar. Then there exist r, s ∈ N, with gcd(r, s) = 1
and distinct parity, such that x = r2 − s2 , y = 2rs, z 2 = r2 + s2 . Note
that r, s, z is again a Pythagorean triple. Now suppose r is even. Then
there exist integers p, q, with gcd(p, q) = 1 and distinct parity, such that
r = 2pq, s = p2 − q 2 , z = p2 + q 2 . So we conclude that
x = (2pq)2 − (p2 − q 2 )2 = −p4 + 6p2 q 2 − q 4 ,
y = 4pq(p4 − q 4 ),
z = p2 + q 2 .
Similarly, when s is even we conclude s = 2pq, r = p2 − q 2 , z = p2 + q 2 and
hence
x = (p2 − q 2 )2 − (2pq)2 = p4 − 6p2 q 2 + q 4 ,
y = 4pq(p2 − q 2 ),
z = p2 + q 2 .
The remaining solutions arise from the previous ones by interchanging x
and y.
(9.8.3) From the previous problem we know that there exist integers p, q such that
abc = ±4pq(p2 − q 2 )(p2 + q 2 )(p4 + p2 q 2 + q 4 − 7p2 q 2 ). Modulo 7 this equals
F.Beukers, Elementary Number Theory
138
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
4pq(p2 − q 2 )(p6 − q 6 ). When 7 divides p or q we are done. When 7 does not
divide pq we have p6 ≡ q 6 ≡ 1(mod 7). Hence p6 − q 6 ≡ 1 − 1 ≡ 0(mod 7),
and we are done again.
(9.8.4) Factorisation of x2 + y 2 = z 3 yields (x + iy)(x − iy) = z 3 . We should
now find the greatest common divisor d of x + iy and x − iy. Note that d
divides the sum 2x and the difference 2iy. Since x, y are relatively prime
we conclude that d divides 2. Hence, up to units in Z[i], d = 1, 1 + i, 2.
Suppose 2 divides x + iy. This implies that x, y are both even, which is
excluded by gcd(x, y) = 1. Notice that 1 + i divides x + iy if and only
if x, y have the same parity, i.e. they are both odd. But then we have
x2 + y 2 ≡ 1 + 1 ≡ 2(mod 4). In other words, x2 + y 2 contains only one
factor 2, so it can never be a square.
We conclude that x + iy and x − iy are relatively prime and hence x + iy
is, up to units, a cube in Z[i]. So there exist a, b ∈ Z such that
x + iy = ϵ(a + bi)3
where ϵ = ±1, ±i. Note that each of these units is a cube, so we can
”absorb” the unit into the cube part. Hence there exist integers a, b such
that x + iy = (a + bi)3 . After comparison of real and imaginary part we
obtain
x = a3 − 3ab2 ,
y = 3a2 b − b3 .
(9.8.6) We start with the equation 2k − 3l = 1. Consider the equation modulo 3.
We see that 2k ≡ 1(mod 3), hence k should be even. From the equation it
follows that 2k − 1 = 3l , hence (2k/2 − 1)(2k/2 + 1) = 3l . Hence the factors
2k/2 ± 1 are either 1 or a power of 3. Write these factors as 3a , 3b with
b < a and note that their difference is 2. I.e. 3a − 3b = 2. In other words,
(3a−b − 1) · 3b = 2 and we conclude that b = 0 and a − b = 1. This implies
that l = a + b = 1 and k = 2. So there are no solutions k, l ≥ 2 in this case.
Now we solve 2k − 3l = −1. Consider the equation modulo 4. We find that
−3l ≡ −1(mod 4). Hence l is even and we can proceed in a similar way
as above. We get 2k = 3l − 1 = (3l/2 − 1)(3l/2 + 1). The factors 3l/2 ± 1
are either 1 or a power of 2. Write the factors as 2a , 2b with b < a. The
difference is 2, so we get 2a − 2b = 2. Hence (2a−b − 1)2b = 2, from which
we conclude that b = 1 and a − b = 1. So k = a + b = 3 and l = 2. This is
the only solution.
NOTE: Catalan’s conjecture has been solved in 2002 by Michailovich.
(9.8.8) Write 4y 2 = x3 + 1 as 4y 2 − 1 = x3 . Factor the left hand side, (2y − 1)(2y +
1) = x3 . The numbers 2y − 1, 2y + 1 are odd and have difference 2. So
they are relatively prime and from (2y + 1)(2y − 1) = x3 it follows that
F.Beukers, Elementary Number Theory
139
2y + 1 = u3 and 2y − 1 = v 3 are cubes. Hence u3 − v 3 = 2. The difference
of two cubes can only be 2 when u = 1, v = −1. One way to see this is to
note that (u − v)(u2 + uv + v 2 ) = 2. Hence u2 + uv + v 2 = ±1, ±2. The
solutions are then found by trying.
So the final solution is y = 0, x = −1.
(9.8.10) We prove the first statement by induction on n. For n = 1 we note that 8
n
n−1
n−1
divides 32 − 1. For larger n we remark that 32 − 1 = (32 − 1)(32 + 1)
n−1
and use the induction hypothesis 2n−1 divides 32
− 1 and the fact that
the second factor is even.
We take ck = 32 , a = 1, b = ck − 1. From the above remark we know that
2k+2 divides bk . Note that
k
N (ak bk ck ) ≤ 3N (bk ) ≤
3
3
b
<
ck .
k
2k+1
2k+1
Hence ck /N (ak , bk , ck ) > 2k+1 /3. The latter tends to ∞ as k → ∞.
(9.8.13) To show this we assume that x3 > y 2 and define δ = x3 − y 2 and We
then apply the abc-conjecture to a = δ/d, b = y 2 /d, c = x3 /d where d =
gcd(x3 , y 2 ). For every ϵ > 0 we get
x3 /d < c(ϵ)rad(x3 y 2 δ/d3 )1+ϵ .
Notice that rad(x3 y 2 δ/d3 ) ≤ xyδ/d. Also notice that y <
So we get
x3 /d < c(ϵ)(xyδ/d)1+ϵ < c(ϵ)(x5/2 δ/d)1+ϵ .
√
x3 − δ < x3/2 .
After multiplication by d1+ϵ we obtain
x3 ≤ x3 dϵ < c(ϵ)(x5/2 δ)1+ϵ .
Now choose ϵ such that −5/2 + 3/(1 + ϵ) = a. Then it follows that
xa < c(ϵ)1/(1+ϵ) δ
from which our assertion follows.
When y 2 > x3 we put δ = y 2 − x3 and find, as above, that
y 2a/3 < c(a)δ.
Noticing that, by assumption, x < y 2/3 , and our assertion follows also in
this case.
F.Beukers, Elementary Number Theory
140
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
(10.3.1) To show part (a) we expand each factor in the product as a geometric series
1
1
1
1
=
1
+
+
+
+ ···
1 − p−s
ps p2s p3s
Taking the product we see that
∏
p≤X,p
1
1 − p−s
prime
equals the sum of n1s over all n which consist entirely of primes ≤ X. This
is certainly larger than the sum of n1s over all n ≤ X.
We should actually have X > 3 in part (b). To show part (b) one uses the
integral criterion
∑ ∫ X−1 dt
>
= log(X − 1)
t
1
n≤X
With a bit more case we can also get the lower bound log(X):
∫ X−1
∑
dt
>1+
= 1 − log(2) + log(X − 1) > log(X).
t
2
n≤X
From (a) and (b) with s = 1 it follows that
∏
p≤X,p
Take logs on both sides
∑
1
> log(X).
−1
1
−
p
prime
− log(1 − p−1 ) > log log(X).
p≤X,p prime
Some calculus shows that − log(1 − x) < x + 4x2 /5 for all x ∈ [0, 1/2].
Hence
∑
4
1
+ 2 > log log(X).
p 5p
p≤X,p prime
Notice also that the sum of 1/p2 over all primes p can be bounded above
by the sum of 1/n2 over all integers n ̸= 1, 4. The latter sum equals π 2 /6 −
1 − 1/16 = 0.58.... Times 4/5 this yields a number < 1/2. So we get
∑
1
1
+
> log log(X)
2 p≤X,p prime p
as desired.
F.Beukers, Elementary Number Theory
141
(10.3.4) Notice that
∏ p2 + 1
=
2−1
p
p prime
∏
p
∏
=
p
=
p2 − 1
(p2 − 1)2
prime
1 − p−4
(1 − p−2 )2
prime
ζ(2)2
(π 2 /6)2
5
= 4
= .
ζ(4)
π /90
2
(10.3.5) Notice that, by definition, n = π(pn ). Hence, by the prime number theorem,
1 pn
= 1.
n→∞ n log(pn )
lim
It now remains to show that
log(pn )
= 1.
n→∞ log(n)
lim
This follows from the fact that for sufficiently large n we have
1 pn
pn
<n<2
,
2 log(pn )
log(pn )
which implies
log(pn ) − log log(pn ) − log(2) < log(n) < log(pn ) − log log(pn ) + log(2).
After division by log(n) and letting n → ∞ we find the desired limit.
(10.3.7) From the previous exercise we know that pn , the n-th prime is asymptotic
√
to n log(n). In particular, for sufficiently large n, pn > n log(n)/2 > n.
So we get
∑
p
∑
1
2
< f inite part +
.
p
log(p)
n
log(n)
·
log(n)/2
prime
n large
The latter infinite series converges by the integral criterion.
Similarly we have for sufficiently large n that pn < 2n log(n) < n2 . Hence
∑ log(p)
∑ 2 log(n)
> f inite part +
.
p
n log(n)/2
p prime
n large
The latter infinite series is the harmonic series which diverges.
F.Beukers, Elementary Number Theory
142
CHAPTER 12. SOLUTIONS TO SELECTED PROBLEMS
(11.4.3) Suppose that e+π and eπ are rational. Then the polynomial (X −e)(X −π)
has rational coefficients. Hence its zeros e, π would be quadratic numbers,
contradicting the fact that e is transcendental.
(11.4.4) Suppose the series has a rational value, say p/q. Choose k and consider the
difference
( )n
p ∑
4
1
k
δ= −
n=0
.
q
5
3n2
2
This is a non-zero rational number with denominator dividing q5k 3k . So
2
we get that δ ≥ 1q 5−k 3−k . On the other hand we have
∞ ( )n
∑
4
1
δ=
.
5
3n2
n=k+1
Let us estimate the terms of this series by (4/5)n 3−(k+1) . Hence
2
δ<
∞
∑
n −(k+1)2
(4/5) 3
−(k+1)2
<3
∞
∑
2
(4/5)n = 5 · 3−(k+1) .
n=0
n=k+1
Comparison of the bounds show that
1 −k −k2
2
5 3
< 5 · 3−(k+1) .
q
Multiplication by 3k gives us 1q 5−k < 5 · 3−2k−1 , hence (9/5)k < 5q/3. This
is impossible if we choose k big enough. Hence our number is irrational.
2
F.Beukers, Elementary Number Theory
Chapter 13
Appendix: Elementary algebra
13.1
Finite abelian groups
In this section we consider some elementary facts on finite abelian groups. The
main application in this course will be to groups of the form (Z/mZ)∗ , the invertible residue classes modulo m. So, if one does not like to work with general
groups one should simply read (Z/mZ)∗ whenever the expression ”finite abelian
group” is used. The order of a finite group is simply the number of elements it
contains.
Lemma 13.1.1 Let G be a finite abelian group of order |G|. Then a|G| = e for
any a ∈ G.
∏
Proof. Let a ∈ G. Consider the product P = g∈G g. Notice that if g runs
through G, then so does ag. Hence
∏
∏
P =
g=
ag.
g∈G
g∈G
The latter product can also be written as a|G|
P = a|G| P and hence e = a|G| .
∏
g∈G
g, which equals a|G| P . So,
2
Remark 13.1.2 Lemma 13.1.1 is actually true for any finite group, also the nonabelian ones. The proof we gave here, due to Lagrange, works only for abelian
groups.
Definition 13.1.3 Let G be a finite group and g ∈ G. The smallest positive
integer k such that g k = e is called the order of g. Notation: ord(g).
Lemma 13.1.4 Let G be a finite group and g ∈ G. Suppose g k = e. Then
ord(g)|k.
143
144
CHAPTER 13. APPENDIX: ELEMENTARY ALGEBRA
Proof. From g k = e follows g k−qord(g) = e for any q ∈ Z. Choose q such that 0 ≤
k − qord(g) < ord(g) and write r = k − qord(g). so g r = e. Since 0 ≤ r < ord(g)
the minimality of ord(g) implies that r = 0 and thus that ord(g) divides k. 2
Corollary 13.1.5 Let G be a finite abelian group. Then ord(g) divides |G| for
any g ∈ G.
Lemma 13.1.6 Let G be a finite abelian group and g, h ∈ G. If ord(g) and
ord(h) are relatively prime then ord(gh) = ord(g)ord(h).
Proof. Let M = ord(gh). From e = (gh)M it follows that e = (gh)M ord(g) =
hM ord(g) . Hence ord(h)|M ord(g). Since (ord(g), ord(h)) = 1 we conclude ord(h)|M .
Similarly, ord(g)|M . Hence ord(g)ord(h)|M . On the other hand, (gh)ord(h)ord(g) =
e and so M |ord(g)ord(h) and thus we find that M = ord(g)ord(h).
2
Lemma 13.1.7 Let G be a finite abelian group and g, h ∈ G. Let L be the lowest
common multiple of the orders ord(g), ord(h). Then there exists k ∈ G such that
L = ord(k).
Proof. Write L as a product of L1 and L2 such that gcd(L1 , L2 ) = 1 and
L1 |ord(g), L2 |ord(h). Notice that g ord(g)/L1 has order L1 . Similarly hord(h)/L2
has order L2 . According to our previous Lemma the product g ord(g)/L1 hord(h)/L2
has order L1 L2 = L.
By repeated application of this Lemma to more than two elements of G we obtain
the following result.
Corollary 13.1.8 Let G be a finite abelian group and let L be the lowest common
multiple of the orders of all elements of G. Then there exists an element in G of
order L.
Definition 13.1.9 Let G be finite group. The annihilator of G is the smallest
positive integer k such that g k = e for all g ∈ G. Notation: Ann(G).
Lemma 13.1.10 Let G be a finite abelian group. Then, Ann(G) = |G| ⇔
G is cyclic.
Proof. The ‘⇐’ being trivial (why?) we shall assume that Ann(G) = |G| and
prove that G is cyclic.
Let L be the lowest common multiple of the orders of all g ∈ G. Notice that
Ann(G) ≤ L ≤ |G|.
The equality Ann(G) = |G| implies |G| = L. According to Lemma 13.1.7 there
exists g ∈ G such that ord(g) = L. Hence, by the equality L = |G|, the element
g generates G and thus G is cyclic.
2
F.Beukers, Elementary Number Theory
13.1. FINITE ABELIAN GROUPS
145
Lemma 13.1.11 Let R be a commutative domain. Then any finite subgroup of
R∗ (the unit group of R) is cyclic.
Proof. Suppose G ⊂ R∗ is a finite subgroup. Let k = Ann(G) and suppose
that |G| > k. Choose g1 , . . . , gk+1 ∈ G distinct. Consider the VanderMonde
determinant
1 1 ...
1 g1 g2 . . . gk+1 δ = ..
.. .
. gk gk . . . gk 1
2
k+1
∏
On the one hand this determinant equals ± i<j (gi − gj ). On the other hand,
since gik = 1 ∀i, we have δ = 0. Since R has no zero divisors this implies that
gi = gj for some i ̸= j, contradicting our choice of distinct elements. So we must
assume that k = |G| and, by Lemma 13.1.10 , G is cyclic.
2
As we said before, the main application of this section is to groups of the form
(Z/mZ)∗ . In particular, Z/pZ is a field when p is prime. Lemma 13.1.11 implies
that (Z/pZ)∗ is a cyclic group. This can be illustrated by the following example
modulo 13. We have
(Z/13Z)∗ = {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
A possible generator of (Z/13Z)∗ is the element 2. Indeed,
(21 , 22 , 23 , 24 , 25 , 26 , 27 , 28 , 29 , 210 , 211 , 212 ) ≡ (2, 4, 8, 3, 6, 12, 11, 9, 5, 10, 7, 1)
and we see that every invertible residue class modulo 13 is a power of 2 modulo
13.
Lemma 13.1.12 Let R be a domain and G a finite subgroup of R∗ . Then,
{
∏
1 if − 1 ̸∈ G
g=
−1 if − 1 ∈ G
g∈G
Proof. Notice that x = 1/x ⇔ x2 = 1 for any x ∈ R. Since R is a domain the
only solutions of x2 = 1 are x = ±1. So, in the product
∏
g
g∈G
all elements cancel except g = 1 and g = −1 if −1 ∈ G. The product of these
exceptional elements is 1 if −1 ̸∈ G and −1 if −1 ∈ G, which proves our lemma. 2
This lemma, applied to R = Z/pZ for any prime p, yields the following result,
(p − 1)! ≡ −1(mod p). This is known as Wilson’s theorem.
F.Beukers, Elementary Number Theory
146
13.2
CHAPTER 13. APPENDIX: ELEMENTARY ALGEBRA
Euclidean domains
Let R be a (not necessarily commutative) domain. The reason to consider non
commutative domains as well is that we would like to include the quaternions in
our considerations. It will not make our statements more difficult, except that we
have to be careful to speak about ‘left-’ and ‘right-’ versions of concepts such as
divisibility. For commutative domains we can drop the suffixes ‘left-’ and ‘right-’.
Definition 13.2.1 A domain R is called (right) euclidean if there exists a function g : R − {0} 7→ N such that
i. g(ab) ≥ g(b), ∀a, b ∈ R − {0}
ii. To every a, b ∈ R with b ̸= 0 there exist q, r ∈ R such that
a = bq + r,
r = 0 or g(r) < g(b).
Definition 13.2.2 Let R be a domain and a, b ∈ R. Then b is called a (right)
divisor of a if there exists c ∈ R such that a = cb.
Theorem 13.2.3 Let R be a (right) euclidean domain and suppose that a1 , . . . , an ∈
R are not all zero. Then there exists a common (right) divisor d ∈ R of a1 . . . , an
such that d = ti ai + · · · tn an for suitable t1 , . . . , tn ∈ R.
Proof. Choose d such that g(d) = min{g(x)| x ∈ S − {0}}, where
S = {x1 a1 + · · · + xn an | x1 , . . . , xn ∈ R}.
Then d is a (right) divisor for any a ∈ R. To see this choose q ∈ R such that
a − qd is either 0 or g(a − qd) < g(d). Since a − qd ∈ S the latter inequality would
contradict the minimality of d. Hence a − qd = 0. In particular d is common
(right) divisor of all ai . Furthermore we know that d ∈ S, so d = t1 a1 + · · · + tn an
for suitable t1 , . . . , tn .
2
As a bonus it follows from Theorem 13.2.3 that any common (right) divisor of the
ai is also a divisor of d. Moreover, by property (i) for euclidean domains we have
g(d) ≥ g(d′ ) for any (right) divisor d′ of d. So d can truly be called a greatest
common divisor of a1 , . . . , an .
Lemma 13.2.4 Let R be a (right) euclidean domain. Suppose g(1) = 1. Then,
ϵ ∈ R∗ ⇐⇒ g(ϵ) = 1.
Proof. ‘⇒’ Choose ϵ′ such that ϵ′ ϵ = 1. Via property (i) of euclidean domains
we see that g(1) ≥ g(ϵ). Since g(1) = 1 and g(ϵ) ∈ N this implies g(ϵ) = 1.
‘⇐’ According to property (ii.) of euclidean domains there exists q ∈ R such that
either 1 − qϵ = 0 or g(1 − qϵ) < g(ϵ). The latter inequality is impossible, hence
we have 1 = qϵ hence ϵ ∈ R∗ .
2
The simplest example of a euclidean domain, Z, is treated in a separate chapter.
In the next three sections we shall deal with three examples which are also of
interest in number theory.
F.Beukers, Elementary Number Theory
13.3. GAUSSIAN INTEGERS
13.3
147
Gaussian integers
Definition 13.3.1 The ring of Gaussian integers is the subring of C given by
{a + bi| a, b ∈ Z}. Notation: Z[i]. The norm of an element α = a + bi ∈ Z[i] is
given by N α := a2 + b2 .
Theorem 13.3.2 We have
a) N αβ = N αN β for any α, β ∈ Z[i]
b) Z[i] is a euclidean domain with the function g(α) = N α
c) The following statements are equivalent,
i. ϵ is a unit in Z[i]
ii. N ϵ = 1
iii. ϵ ∈ {1, −1, i, −i}.
Proof. a) Note that N α = |α|2 where | · | is the ordinary absolute value on C.
Our assertion follows from |αβ| = |α| · |β|.
b) Notice that N α = 0 ⇔ α = 0 and N α ≥ 1 for all α ̸= 0. Property (i) for
euclidean domains now follows from a). To show property (ii) we work momentarily in C. Let α/β = x + yi ∈ C. Let κ = p + qi where p, q are the nearest
integers to x and y respectively. So,
|(x + yi − κ)|2 ≤ (1/2)2 + (1/2)2 = 1/2.
Hence, after multiplication by N β, N (α − κβ) ≤ N β/2. This proves property
(ii) for euclidean domains.
c) Equivalence of i. and ii. follows from Lemma 13.2.4 and the fact that N 1 = 1.
Equivalence of ii. and iii. follows from the observation
a2 + b2 = 1 ⇔ (a, b) ∈ {(1, 0), (−1, 0), (0, 1), (0, −1)}.
2
13.4
Quaternion integers
First of all we recall the definition of quaternions.
Definition 13.4.1 The quaternions are expressions of the form a + bi + cj +
dk, a, b, c, d ∈ R with addition given by
(a + bi + cj + dk) + (a′ + b′ i + c′ j + d′ k) = (a + a′ ) + (b + b′ )i + (c + c′ )j + (d + d′ )k
F.Beukers, Elementary Number Theory
148
CHAPTER 13. APPENDIX: ELEMENTARY ALGEBRA
and multiplication generated by the rules
i2
ij
jk
ki
= j2 =
= k,
= i,
= j,
k2
ji
kj
ik
= −1,
= −k,
= −i,
= −j.
Notation: H
With the above addition and multiplication the quaternions form a (non-commutative)
domain. The conjugate of a quaternion α = a+bi+cj+dk is given by a−bi−cj−dk
and denoted by α. The norm of a quaternion α = a + bi + cj + dk is given by
N α = αα and equals a2 + b2 + c2 + d2 (verify!). This implies that the inverse of
a non-zero quaternion α exists and is given by α/N α. Furthermore,
Theorem 13.4.2 Let α and β be quaternions. Then N αβ = N α · N β.
Proof. Write α = a + bi + cj + dk and β = a′ + b′ i + c′ j + d′ k. Then,
(a + bi + cj + dk)(a′ + b′ i + c′ j + d′ k) =
= (aa′ − bb′ − cc′ − dd′ )
+(ab′ + ba′ + cd′ − dc′ )i
+(ac′ − bd′ + ca′ + db′ )j
+(ad′ + bc′ − cb′ + da′ )k
Our statement now follows from Euler’s identity,
(a2 + b2 + c2 + d2 )(a′2 + b′2 + c′2 + d′2 ) =
= (aa′ − bb′ − cc′ − dd′ )2
+(ab′ + ba′ + cd′ − dc′ )2
+(ac′ − bd′ + ca′ + db′ )2
+(ad′ + bc′ − cb′ + da′ )2 .
2
An alternative way to describe quaternions is to identify a + bi + cj + dk with
the matrix
(
)
a + bi c + di
−c + di a − bi
√
where i = −1. Addition and multiplication of quaternions comes down to addition and multiplication of the corresponding matrices. The matrix corresponding
to the conjugate reads
(
)
a − bi −c − di
.
c − di a + bi
F.Beukers, Elementary Number Theory
13.4. QUATERNION INTEGERS
149
The norm is simply the determinant of the corresponding matrix. Associativity
of quaternion multiplication now follows directly from associativity of matrix
multiplication. Also, the relation αα = a2 + b2 + c2 + d2 is now easy to verify.
Definition 13.4.3 Let notations be as above. The quaternion integers are given
by the set
1
{ (p + qi + rj + sk)| p, q, r, s ∈ Z, p ≡ q ≡ r ≡ s(mod 2)}.
2
Notation: q.
We leave it to the reader to verify that q is a subring of H and that N α ∈ N for
all α ∈ q, α ̸= 0.
Theorem 13.4.4 We have
a) q is a (right-)euclidean domain with the function g(α) = N α.
b) The following statements are equivalent,
i. ϵ is a unit in q
ii. N ϵ = 1
iii. ϵ ∈ {±1, ±i, ±j, ±k, 12 (±1 ± i ± j ± k)}.
c) To any quaternionic integer α there exists a unit ϵ such that αϵ is of the
form p + qi + rj + sk, p, q, r, s ∈ Z.
Proof. a) Notice that N α = 0 ⇔ α = 0 and N α ≥ 1 for all α ̸= 0. Property
(i) for euclidean domains now follows from Theorem 13.4.2. To show property
(ii) we work momentarily in H. Let αβ −1 = x + yi + zj + uk ∈ H. Let κ =
p + qi + rj + sk where p, q, r, s are the nearest integers to x, y, z, u respectively.
So, N (x + yi + zj + uk − κ) ≤ (1/2)2 + (1/2)2 + (1/2)2 + (1/2)2 = 1. Hence, after
multiplication by N β, N (α − κβ) ≤ N β. This proves property (ii) for euclidean
domains when strict inequality holds. Notice that the equality sign holds if and
only if all four numbers x, y, z, u are halfintegral. But then we have automatically
x + yi + zj + uk ∈ q. So property (ii) holds in all cases.
b) Equivalence of i. and ii. follows from Lemma 13.2.4 and the fact that N 1 = 1.
Equivalence of ii. and iii. follows by determination of all integral solutions of
a2 + b2 + c2 + d2 = 4, a ≡ b ≡ c ≡ d(mod 2).
c) If α is already of the form p + qi + rj + sk, p, q, r, s ∈ Z we can take ϵ = 1. If
α is of the form (p + qi + rj + sk)/2, p, q, r, s ∈ Zodd we write p = 4p1 + p2 , q =
4q1 + q2 , r = 4r1 + r2 , s = 4s1 + s2 with p2 , q2 , r2 , s2 ∈ {±1}. Then
α = 2(p1 + q1 i + r1 j + s1 k) + (p2 + q2 i + r2 j + s2 k)/2.
Observe that we can now take ϵ = (p2 − q2 i − r2 j − s2 k)/2
2
It is a nice exercise to show that the unit group in q modulo ±1 is isomorphic to
the alternating group A4 .
F.Beukers, Elementary Number Theory
150
13.5
CHAPTER 13. APPENDIX: ELEMENTARY ALGEBRA
Polynomials
Let F be a field and F [X] the corresponding polynomial ring in one variable. As
well known, any non-zero polynomial f ∈ F [X] has a degree, which we denote by
deg(f ). We have deg(f g) = deg(f ) + deg(g) and deg(f + g) ≤ deg(f ) + deg(g).
Theorem 13.5.1 Let F [X] be a polynomial ring over a field F . Then,
a) F [X] is a euclidean domain with the function g(f ) = 2deg(f ) if f ̸≡ 0.
b) f ∈ F [X]∗ if and only if f is a constant non-zero polynomial.
Proof. Property (a) follows from the well-known division algorithm for polynomials. Property (b) is a consequence of Lemma 13.2.4
F.Beukers, Elementary Number Theory
Index
amicable numbers, 17
Bertrand’s postulate, 110
Cantor diogonal, 121
Carmichael numbers, 34
character, 62
circle method, 78
conductor, 100
congruent, 21
congruent number, 95
conjugate, 85
continued fraction algorithm, 82
convergents, 83
critical strip, 109
decryption, 44
descent, 97
diophantine equation, 94
Dirichlet character, 64
discriminant, 85
divisor sum, 15
encryption, 44
euclidean algorithm, 9
Euler product, 109
even character, 65
Fermat numbers, 37, 54
Gauss sum, 65
greatest common divisor, 7
Hilbert tenth problem, 94
invertible, 21
irrational, 117
Jacobi sum, 66
Jacobi symbol, 56
Jacobsthal sum, 69
lcm, 11
Legendre symbol, 47
lowest common multiple, 11
Mersenne numbers, 17, 37
Mersenne prime, 17
minimal period, 28
minimal polynomial, 85
Mordell equation, 99
negative resdue class, 49
non-residue, 47
odd character, 65
partial fractions, 83
perfect number, 17
period, 28
periodic, 28
periodic continued fraction, 86
Pollard rho, 40
positive residue class, 49
primitive root, 25
principal character, 62
public key, 44
purely periodic, 28, 86
Pythagorean triplet, 95
quadratic
quadratic
quadratic
quadratic
irrational, 85
non-residue, 47
reciprocity, 48
residue, 47
radical, 100
reciprocity law, 67
151
152
reduced quadratic irrational, 85
relatively prime, 7
repunits, 53
residue class, 21
Riemann hypothesis, 109
secret key, 44
sieve methods, 111
terminate, 82
totient function, 24
transcendental, 120
trivial character, 62
twin prime, 110
Waring’s problem, 78
witness, 35
zer-knowledge proofs, 45
F.Beukers, Elementary Number Theory
INDEX